1. A few analysis questions

Hey everyone! Any input would be appreciated.

A note on notation, $\displaystyle f\in\mathfrak{R}(\alpha)$ on $\displaystyle [a,b]$ means that $\displaystyle f$ is Riemann-Stieltjes integrable with respect to $\displaystyle \alpha(x)$ on the interval $\displaystyle [a,b]$

Question 1: Let $\displaystyle x_0\in[a,b]$ and define $\displaystyle f$ on $\displaystyle [a,b]$ by $\displaystyle f:x\longmapsto\left\{ \begin{array}{rcl} 0 & \mbox{if} & x\ne x_0\\ c & \mbox{if} & x=x_0 \end{array} \right.$. Also, let $\displaystyle \alpha(x)$ be continuous at $\displaystyle x=x_0$. Is$\displaystyle f\in\mathfrak{R}(\alpha)$ on $\displaystyle [a,b]$? If so what is its value.

Lemma: Suppose that $\displaystyle f$ is bounded on $\displaystyle [a,b]$, $\displaystyle f$ has a finite number of discontinuities and $\displaystyle \alpha$ is continuous at all those points, then $\displaystyle f\in\mathfrak{R}(\alpha)$ on $\displaystyle [a,b]$.

We can see that $\displaystyle f\in\mathfrak{R}(\alpha)$ since $\displaystyle f$ has only one discontinuity, at $\displaystyle x_0$, and $\displaystyle \alpha$ is continuous there.

Next to find the value of this integral consider $\displaystyle L\left(P,f,\alpha\right)$ for any partitions. Now consider the interval $\displaystyle [x_{i-1},x_i]$ created by $\displaystyle P$. Since $\displaystyle P$ is a finite point set, it follows that $\displaystyle [x_{i-1},x_i]$ has infinite points. Which indicates it contains values other than $\displaystyle x_0$. So for any interval on $\displaystyle P$ we can see that $\displaystyle m_i=\inf_{[x_{i-1},x_i]}f=0$. Now since $\displaystyle L\left(P,f,\alpha\right)=\sum_{i=1}^{n}m_i\Delta \alpha_i$ we can see that for any partition $\displaystyle L\left(P,f,\alpha\right)=0$. Now since $\displaystyle f\in\mathfrak{R}(\alpha)$ it follows that $\displaystyle \int_a^b f~d\alpha=\inf_{P}L\left(P,f,\alpha\right)=0$
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Question 2: Let $\displaystyle f:x\longmapsto \left\{ \begin{array}{rcl} 0 & \mbox{if} & x\not\in\mathbb{Q}\\ 1 & \mbox{if} & x\in\mathbb{Q} \end{array} \right.$, this is known as Dirchlet's function. Show that $\displaystyle f\not\in\mathfrak{R}(\alpha)$ for any $\displaystyle a<b$

Answer: This one is simple I think. Consider any partition $\displaystyle P$ of $\displaystyle [a,b]$. Now consider any interval $\displaystyle [x_{i-1},x_i]$ formed by $\displaystyle P$. For the same reason as above $\displaystyle [x_{i-1},x_i]$ is infinite. Now since $\displaystyle R-\left\{\mathbb{Q}\right\}$ is dense in any subset of the reals it follows that there are value of $\displaystyle x$ in $\displaystyle [x_{i-1},x_i]$ such that $\displaystyle x\not\in\mathbb{Q}$ where it follows similarly to above that $\displaystyle L\left(P,f,\alpha\right)=0$ for any partition. Now following a similar argument noting that $\displaystyle \mathbb{Q}$ is dense in any subset of the reals we can see that for any partition $\displaystyle U\left(P,f,\alpha\right)=1$. So for any partition $\displaystyle U\left(P,f,\alpha\right)-L\left(P,f,\alpha\right)=1$. Where it follows that $\displaystyle f\not\in\mathfrak{R}(\alpha)$ since it fails the criterion that given any $\displaystyle \varepsilon>0$ we must have a partition such that $\displaystyle U\left(P,f,\alpha\right)-L\left(P,f,\alpha\right)<\varepsilon$

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Question 3: Does $\displaystyle f^2\in\mathfrak{R}(\alpha)\implies f\in\mathfrak{R}(\alpha)$?

Answer: No. Counterexample $\displaystyle f:x\longmapsto\left\{ \begin{array}{rcl} -1 & \mbox{if} & x\not\in\mathbb{Q}\\ 1 & \mbox{if} & x\in\mathbb{Q}\end{array}\right.$. The reason is that $\displaystyle f^2:x\longmapsto\left\{ \begin{array}{rcl} 1 & \mbox{if} & x\not\in\mathbb{Q}\\ 1 & \mbox{if} & x\in\mathbb{Q}\end{array} \right.=1$ which is continuous thus integrable, but for the same reason as $\displaystyle f$ in question 2 $\displaystyle f$ is not integrable.

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Question 4: Suppose that $\displaystyle f\geqslant 0$, $\displaystyle f$ is continuous on $\displaystyle [a,b]$ and $\displaystyle \int_a^b f~dx=0$. Prove that $\displaystyle f(x)=0~~\forall x\in[a,b]$.

Answer: It is obvious that if $\displaystyle f=0\implies \int_a^b f~dx=0$. So instead let us prove that if $\displaystyle f>0\implies \int_a^b f~dx>0$.

Since $\displaystyle f\ne 0$ we can find a Partition of $\displaystyle [a,b]$ such that for some interval $\displaystyle [x_{i-1},x_i]$ we have that $\displaystyle f>0~~\forall x \in[x_{i-1},x_i]$, now since this interval is closed it follows that $\displaystyle \inf_{[x_{i-1},x_i]}f>0$. Because of this $\displaystyle L\left(P,f,\right)>0$. Now consider that since $\displaystyle f\in\mathfrak{R}$ it follows that $\displaystyle \int_a^b f~dx=\sup_{P}L\left(P,f\right)>0$.

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Question 5: Prove that if $\displaystyle f\in\mathfrak{R}$ on $\displaystyle [a,b]$ and $\displaystyle c\in[a,b]$ that $\displaystyle \int_a^c f~d\alpha+\int_c^b f~d\alpha=\int_a^b f~d\alpha$

Answer: Let $\displaystyle P_1,P_2$ be partitions of $\displaystyle [a,c],[b,c]$ respectively. Let $\displaystyle P=P_1\cup P_2$.

Let us note that $\displaystyle L\left(P,f,\alpha\right)\leq L\left(P_1,f,\alpha\right)+L\left(P_2,f,\alpha\rig ht)$. To prove this simply note that given an interval $\displaystyle [x_{i-1},x_i]$ and $\displaystyle x\in[x_{i-1},x_i]$ that $\displaystyle (x_{i}-x_{i-1})\inf_{[x_{i-1},x_i]}f\leqslant (x-x_{i-1})\inf_{[x_{i-1},x]}f+(x_i-x)\inf_{[x,x_i]}f$.

So since $\displaystyle L\left(P,f,\alpha\right)\leqslant \int_a^c f~d\alpha+\int_c^b f~d\alpha$ taking the infimum over both intervals gives $\displaystyle \int_a^b f~d\alpha\leqslant \int_a^c f~d\alpha+\int_c^b f~d\alpha\quad{\color{red}(1)}$.

Using the same argument except considering that $\displaystyle U\left(P_1,f,f\alpha\right)+U\left(P_2,f,\alpha\ri ght)\leqslant U\left(P,f,\alpha\right)$ we arrive at $\displaystyle \int_a^c f~d\alpha+\int_c^b f~d\alpha\leqslant \int_a^b f~d\alpha\quad{\color{red}(2)}$

Combining $\displaystyle \color{red}(1),(2)$ completes the proof.

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Question six: Prove Holder's inequality for integrals, i.e. if $\displaystyle \frac{1}{p}+\frac{1}{q}\leqslant 1$ then $\displaystyle \left|\int_a^b fg~d\alpha\right|\leqslant \left\{\int_a^b f^p~d\alpha\right\}^{\frac{1}{p}}\cdot\left\{\int_ a^b |g|^q~d\alpha\right\}^{\frac{1}{q}}$

I can prove this, but it requires me first to prove that if the conditions of $\displaystyle p,q$ are met as above then $\displaystyle uv\leq\frac{u^p}{p}+\frac{v^q}{q}$ and I am not sure how to do this.

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For 1-5 any criticism is appreciated ...for number six any help with the inequality would be great.

2. Originally Posted by Mathstud28

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Question 4: Suppose that $\displaystyle f\geqslant 0$, $\displaystyle f$ is continuous on $\displaystyle [a,b]$ and $\displaystyle \int_a^b f~dx=0$. Prove that $\displaystyle f(x)=0~~\forall x\in[a,b]$.

Answer: It is obvious that if $\displaystyle f=0\implies \int_a^b f~dx=0$. So instead let us prove that if $\displaystyle f>0\implies \int_a^b f~dx>0$.

Since $\displaystyle f\ne 0$ we can find a Partition of $\displaystyle [a,b]$ such that for some interval $\displaystyle [x_{i-1},x_i]$ we have that $\displaystyle f>0~~\forall x \in[x_{i-1},x_i]$, now since this interval is closed it follows that $\displaystyle \inf_{[x_{i-1},x_i]}f>0$. Because of this $\displaystyle L\left(P,f,\right)>0$. Now consider that since $\displaystyle f\in\mathfrak{R}$ it follows that $\displaystyle \int_a^b f~dx=\sup_{P}L\left(P,f\right)>0$.

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For 1-5 any criticism is appreciated ...for number six any help with the inequality would be great.

hmm, i just wonder about item 4 because i really can't comprehend on the way you did it..
why prove this:
Originally Posted by Mathstud28
let us prove that if $\displaystyle f>0\implies \int_a^b f~dx>0$
whence you want to prove this:
Originally Posted by Mathstud28
Suppose that $\displaystyle f\geqslant 0$, $\displaystyle f$ is continuous on $\displaystyle [a,b]$ and $\displaystyle \int_a^b f~dx=0$. Prove that $\displaystyle f(x)=0~~\forall x\in[a,b]$

3. Thank you Kalagota. Let me rephrase this a little better

Overview: Before I do my proof let me explain what I attempt to do. I will first show that $\displaystyle \int_a^b 0~d\alpha=0$. I will then show that if $\displaystyle f>0$ at any point of $\displaystyle [a,b]$ then $\displaystyle \int_a^b f~d\alpha>0$. This will show that the only function such that $\displaystyle \int_a^b f~d\alpha=0$ with $\displaystyle f\geqslant 0$ is $\displaystyle f:[a,b]\longmapsto 0$.

Part one: Here I will show that $\displaystyle \int_a^b f~d\alpha=0$. Consider a partition $\displaystyle P$ of $\displaystyle [a,b]$. Let $\displaystyle [x_{i-1},x_i]$ be any interval formed by $\displaystyle P$. It is clear that since $\displaystyle f:[x_{i-1},x_i]\longmapsto0$ that $\displaystyle M_i=\sup_{[x_{i-1},x_i]}f=0$. So $\displaystyle U\left(P,f,\alpha\right)=\sum_{i=1}^{n}m_i\cdot \Delta\alpha_i=0$. So since $\displaystyle f:[a,b]\longmapsto0$ is continuous it follows that $\displaystyle \int_a^b 0~d\alpha=\inf_{P}~U\left(P,f,\alpha\right)=0$

Part two: Let $\displaystyle f>0$ for some point of $\displaystyle [a,b]$. We know use the following lemma.

Lemma: I will show that if $\displaystyle f$ is uniformly continuous and $\displaystyle f(c)>0$ then there exists a neighborhood of $\displaystyle c$ such that $\displaystyle f>0$.

Proof: Since $\displaystyle f$ is uniformly continuous we have that for every $\displaystyle \varepsilon>0$ there exists a corresponding $\displaystyle \delta>0$ such that $\displaystyle |x-y|<\delta\implies|f(x)-f(y)|<\varepsilon$. Now let $\displaystyle f(c)>0$, this implies that $\displaystyle |c-y|<\delta\implies |f(c)-f(y)|<\varepsilon$. So assume that there does not exist a neighborhood of $\displaystyle c$ such that $\displaystyle f(y)>0$, this implies that $\displaystyle f(c)<|f(c)-f(y)|$. So picking $\displaystyle \varepsilon=f(c)$ we cannot a $\displaystyle \delta$ that satisfies $\displaystyle |f(c)-f(y|<\varepsilon$ which violates continuity.

So now noticing that $\displaystyle f$ being on continuous on $\displaystyle [a,b]$ implies that $\displaystyle f$ is uniformly continuous on $\displaystyle [a,b]$ since $\displaystyle [a,b]$ is compact, we can see that there exists a $\displaystyle [d,e]\subset [a,b]$ such that $\displaystyle f([d,e])>0$. So now pick a partition of $\displaystyle P$ such that $\displaystyle d,e\in P$. It is clear that there is an interval $\displaystyle [x_{i-1},x_i]\subset [d,e]$. So we can see that $\displaystyle m_i=\inf_{[x_{i-1},x_i]}f>0$ which implies that $\displaystyle L\left(P,f,\alpha\right)>0$. So now since $\displaystyle f$ is continuous we can see that $\displaystyle \int_a^b f~d\alpha=\sup_{P}L\left(P,f,\alpha\right)>0$.

This proves that the only continuous function $\displaystyle f$ on $\displaystyle [a,b]$ such that $\displaystyle \int_a^b f~d\alpha=0$ is $\displaystyle f:[a,b]\longmapsto0\quad\blacksquare$

4. These all look correct to me. One small point on Q.2:

Originally Posted by Mathstud28
Now following a similar argument noting that $\displaystyle \mathbb{Q}$ is dense in any subset of the reals we can see that for any partition $\displaystyle U\left(P,f,\alpha\right)=1$.
The supremum of the function in any interval of the partition is 1, but the upper sum U for the partition is $\displaystyle \alpha(b)-\alpha(a)$.

Originally Posted by Mathstud28
Question six: Prove Holder's inequality for integrals, i.e. if $\displaystyle \frac{1}{p}+\frac{1}{q}\leqslant 1$ then $\displaystyle \left|\int_a^b fg~d\alpha\right|\leqslant \left\{\int_a^b |f|^p~d\alpha\right\}^{\frac{1}{p}}\cdot\left\{\in t_a^b |g|^q~d\alpha\right\}^{\frac{1}{q}}$
Not true. You need $\displaystyle \tfrac1p + \tfrac1q\; {\color{red}=}\; 1$ (or possibly $\displaystyle {\color{red}\geqslant}\;1$).

Originally Posted by Mathstud28
it requires me first to prove that if the conditions of $\displaystyle p,q$ are met as above then $\displaystyle uv\leq\frac{u^p}{p}+\frac{v^q}{q}$ and I am not sure how to do this.
Assuming that $\displaystyle \tfrac1p + \tfrac1q = 1$ and that u and v are ≥0, this is elementary calculus. Let $\displaystyle f(u,v) = \frac{u^p}{p}+\frac{v^q}{q} - uv$. For a fixed value of v, f(u,v) is ≥0 at u=0 and as u→∞. The derivative is 0 when $\displaystyle v = u^{p-1}$, at which point f has a minimum value of 0.

5. The supremum of the function in any interval of the partition is 1, but the upper sum U for the partition is $\displaystyle \alpha(b)-\alpha(a)$.
Yes, thank you. I had this correct down on my paper

$\displaystyle U\left(P,f,\alpha\right)=\sum_{i=1}^{n}M_i\cdot\De lta\alpha_i=\sum_{i=1}^{n}\Delta \alpha_i$$\displaystyle =\left(\alpha(x_1)-\alpha(x_0)\right)+(\alpha(x_2)-\alpha(x_1))+\cdots+(\alpha(x_n)-\alpha(x_{n-1})=\alpha(x_n)-\alpha(x_0)$$\displaystyle =\alpha(b)-\alpha(a)$

Not true. You need $\displaystyle \tfrac1p + \tfrac1q\; {\color{red}=}\; 1$ (or possibly $\displaystyle {\color{red}\geqslant}\;1$).[/tex]
Pardon me, I transcribed the problem incorrectly.

Assuming that $\displaystyle \tfrac1p + \tfrac1q = 1$ and that u and v are ≥0, this is elementary calculus. Let $\displaystyle f(u,v) = \frac{u^p}{p}+\frac{v^q}{q} - uv$. For a fixed value of v, f(u,v) is ≥0 at u=0 and as u→∞. The derivative is 0 when $\displaystyle v = u^{p-1}$, at which point f has a minimum value of 0.
I meant using purely non-calculus means, but I guess this works as well

6. Originally Posted by Mathstud28
Hey everyone! Any input would be appreciated.

A note on notation, $\displaystyle f\in\mathfrak{R}(\alpha)$ on $\displaystyle [a,b]$ means that $\displaystyle f$ is Riemann-Stieltjes integrable with respect to $\displaystyle \alpha(x)$ on the interval $\displaystyle [a,b]$

Question 1: Let $\displaystyle x_0\in[a,b]$ and define $\displaystyle f$ on $\displaystyle [a,b]$ by $\displaystyle f:x\longmapsto\left\{ \begin{array}{rcl} 0 & \mbox{if} & x\ne x_0\\ c & \mbox{if} & x=x_0 \end{array} \right.$. Also, let $\displaystyle \alpha(x)$ be continuous at $\displaystyle x=x_0$. Is$\displaystyle f\in\mathfrak{R}(\alpha)$ on $\displaystyle [a,b]$? If so what is its value.

Lemma: Suppose that $\displaystyle f$ is bounded on $\displaystyle [a,b]$, $\displaystyle f$ has a finite number of discontinuities and $\displaystyle \alpha$ is continuous at all those points, then $\displaystyle f\in\mathfrak{R}(\alpha)$ on $\displaystyle [a,b]$.

We can see that $\displaystyle f\in\mathfrak{R}(\alpha)$ since $\displaystyle f$ has only one discontinuity, at $\displaystyle x_0$, and $\displaystyle \alpha$ is continuous there.

Next to find the value of this integral consider $\displaystyle L\left(P,f,\alpha\right)$ for any partitions. Now consider the interval $\displaystyle [x_{i-1},x_i]$ created by $\displaystyle P$. Since $\displaystyle P$ is a finite point set, it follows that $\displaystyle [x_{i-1},x_i]$ has infinite points. Which indicates it contains values other than $\displaystyle x_0$. So for any interval on $\displaystyle P$ we can see that $\displaystyle m_i=\inf_{[x_{i-1},x_i]}f=0$. Now since $\displaystyle L\left(P,f,\alpha\right)=\sum_{i=1}^{n}m_i\Delta \alpha_i$ we can see that for any partition $\displaystyle L\left(P,f,\alpha\right)=0$. Now since $\displaystyle f\in\mathfrak{R}(\alpha)$ it follows that $\displaystyle \int_a^b f~d\alpha=\inf_{P}L\left(P,f,\alpha\right)=0$
Looks good to me.

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Question 2: Let $\displaystyle f:x\longmapsto \left\{ \begin{array}{rcl} 0 & \mbox{if} & x\not\in\mathbb{Q}\\ 1 & \mbox{if} & x\in\mathbb{Q} \end{array} \right.$, this is known as Dirchlet's function. Show that $\displaystyle f\not\in\mathfrak{R}(\alpha)$ for any $\displaystyle a<b$

Answer: This one is simple I think. Consider any partition $\displaystyle P$ of $\displaystyle [a,b]$. Now consider any interval $\displaystyle [x_{i-1},x_i]$ formed by $\displaystyle P$. For the same reason as above $\displaystyle [x_{i-1},x_i]$ is infinite. Now since $\displaystyle R-\left\{\mathbb{Q}\right\}$ is dense in any subset of the reals it follows that there are value of $\displaystyle x$ in $\displaystyle [x_{i-1},x_i]$ such that $\displaystyle x\not\in\mathbb{Q}$ where it follows similarly to above that $\displaystyle L\left(P,f,\alpha\right)=0$ for any partition. Now following a similar argument noting that $\displaystyle \mathbb{Q}$ is dense in any subset of the reals we can see that for any partition $\displaystyle U\left(P,f,\alpha\right)=1$. So for any partition $\displaystyle U\left(P,f,\alpha\right)-L\left(P,f,\alpha\right)=1$. Where it follows that $\displaystyle f\not\in\mathfrak{R}(\alpha)$ since it fails the criterion that given any $\displaystyle \varepsilon>0$ we must have a partition such that $\displaystyle U\left(P,f,\alpha\right)-L\left(P,f,\alpha\right)<\varepsilon$
Also good.
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Question 3: Does $\displaystyle f^2\in\mathfrak{R}(\alpha)\implies f\in\mathfrak{R}(\alpha)$?

Answer: No. Counterexample $\displaystyle f:x\longmapsto\left\{ \begin{array}{rcl} -1 & \mbox{if} & x\not\in\mathbb{Q}\\ 1 & \mbox{if} & x\in\mathbb{Q}\end{array}\right.$. The reason is that $\displaystyle f^2:x\longmapsto\left\{ \begin{array}{rcl} 1 & \mbox{if} & x\not\in\mathbb{Q}\\ 1 & \mbox{if} & x\in\mathbb{Q}\end{array} \right.=1$ which is continuous thus integrable, but for the same reason as $\displaystyle f$ in question 2 $\displaystyle f$ is not integrable.
Good.

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Question 4: Suppose that $\displaystyle f\geqslant 0$, $\displaystyle f$ is continuous on $\displaystyle [a,b]$ and $\displaystyle \int_a^b f~dx=0$. Prove that $\displaystyle f(x)=0~~\forall x\in[a,b]$.

Answer: It is obvious that if $\displaystyle f=0\implies \int_a^b f~dx=0$.
Yes, it is obvious. You were also not asked to prove it so there is no need to mention it! (This is NOT an "if and only if" statement.)

So instead let us prove that if $\displaystyle f>0\implies \int_a^b f~dx>0$.
Since $\displaystyle f\ne 0$ we can find a Partition of $\displaystyle [a,b]$ such that for some interval $\displaystyle [x_{i-1},x_i]$ we have that $\displaystyle f>0~~\forall x \in[x_{i-1},x_i]$, now since this interval is closed it follows that $\displaystyle \inf_{[x_{i-1},x_i]}f>0$. Because of this $\displaystyle L\left(P,f,\right)>0$. Now consider that since $\displaystyle f\in\mathfrak{R}$ it follows that $\displaystyle \int_a^b f~dx=\sup_{P}L\left(P,f\right)>0$.
Good.
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Question 5: Prove that if $\displaystyle f\in\mathfrak{R}$ on $\displaystyle [a,b]$ and $\displaystyle c\in[a,b]$ that $\displaystyle \int_a^c f~d\alpha+\int_c^b f~d\alpha=\int_a^b f~d\alpha$

Answer: Let $\displaystyle P_1,P_2$ be partitions of $\displaystyle [a,c],[b,c]$ respectively. Let $\displaystyle P=P_1\cup P_2$.

Let us note that $\displaystyle L\left(P,f,\alpha\right)\leq L\left(P_1,f,\alpha\right)+L\left(P_2,f,\alpha\rig ht)$. To prove this simply note that given an interval $\displaystyle [x_{i-1},x_i]$ and $\displaystyle x\in[x_{i-1},x_i]$ that $\displaystyle (x_{i}-x_{i-1})\inf_{[x_{i-1},x_i]}f\leqslant (x-x_{i-1})\inf_{[x_{i-1},x]}f+(x_i-x)\inf_{[x,x_i]}f$.

So since $\displaystyle L\left(P,f,\alpha\right)\leqslant \int_a^c f~d\alpha+\int_c^b f~d\alpha$ taking the infimum over both intervals gives $\displaystyle \int_a^b f~d\alpha\leqslant \int_a^c f~d\alpha+\int_c^b f~d\alpha\quad{\color{red}(1)}$.

Using the same argument except considering that $\displaystyle U\left(P_1,f,f\alpha\right)+U\left(P_2,f,\alpha\ri ght)\leqslant U\left(P,f,\alpha\right)$ we arrive at $\displaystyle \int_a^c f~d\alpha+\int_c^b f~d\alpha\leqslant \int_a^b f~d\alpha\quad{\color{red}(2)}$

Combining $\displaystyle \color{red}(1),(2)$ completes the proof.
I think you need to say a little more than this. Not every partition of [a, b] can be writtenas $\displaystyle P_1\cup P_2$ in this way. You need to assert that given any such partition, there exist a refinement that is of that form. If c is not itself a "boundary" of an interval in P, it lies in one of the intervals. Dividing that interval at c gives a refinement that can be written as $\displaystyle P_1\cup P_2$.

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Question six: Prove Holder's inequality for integrals, i.e. if $\displaystyle \frac{1}{p}+\frac{1}{q}\leqslant 1$ then $\displaystyle \left|\int_a^b fg~d\alpha\right|\leqslant \left\{\int_a^b f^p~d\alpha\right\}^{\frac{1}{p}}\cdot\left\{\int_ a^b |g|^q~d\alpha\right\}^{\frac{1}{q}}$

I can prove this, but it requires me first to prove that if the conditions of $\displaystyle p,q$ are met as above then $\displaystyle uv\leq\frac{u^p}{p}+\frac{v^q}{q}$ and I am not sure how to do this.
That is called "Young's inequality" and a proof can be found here:
Young's inequality - Wikipedia, the free encyclopedia

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For 1-5 any criticism is appreciated ...for number six any help with the inequality would be great.

7. Originally Posted by HallsofIvy
I think you need to say a little more than this. Not every partition of [a, b] can be writtenas $\displaystyle P_1\cup P_2$ in this way. You need to assert that given any such partition, there exist a refinement that is of that form. If c is not itself a "boundary" of an interval in P, it lies in one of the intervals. Dividing that interval at c gives a refinement that can be written as $\displaystyle P_1\cup P_2$.
Thank you HallsofIvy.

8. Originally Posted by Mathstud28
Question 1: Let $\displaystyle x_0\in[a,b]$ and define $\displaystyle f$ on $\displaystyle [a,b]$ by $\displaystyle f:x\longmapsto\left\{ \begin{array}{rcl} 0 & \mbox{if} & x\ne x_0\\ c & \mbox{if} & x=x_0 \end{array} \right.$. Also, let $\displaystyle \alpha(x)$ be continuous at $\displaystyle x=x_0$. Is$\displaystyle f\in\mathfrak{R}(\alpha)$ on $\displaystyle [a,b]$? If so what is its value.

Lemma: Suppose that $\displaystyle f$ is bounded on $\displaystyle [a,b]$, $\displaystyle f$ has a finite number of discontinuities and $\displaystyle \alpha$ is continuous at all those points, then $\displaystyle f\in\mathfrak{R}(\alpha)$ on $\displaystyle [a,b]$.

We can see that $\displaystyle f\in\mathfrak{R}(\alpha)$ since $\displaystyle f$ has only one discontinuity, at $\displaystyle x_0$, and $\displaystyle \alpha$ is continuous there.

Next to find the value of this integral consider $\displaystyle L\left(P,f,\alpha\right)$ for any partitions. Now consider the interval $\displaystyle [x_{i-1},x_i]$ created by $\displaystyle P$. Since $\displaystyle P$ is a finite point set, it follows that $\displaystyle [x_{i-1},x_i]$ has infinite points. Which indicates it contains values other than $\displaystyle x_0$. So for any interval on $\displaystyle P$ we can see that $\displaystyle m_i=\inf_{[x_{i-1},x_i]}f=0$. Now since $\displaystyle L\left(P,f,\alpha\right)=\sum_{i=1}^{n}m_i\Delta \alpha_i$ we can see that for any partition $\displaystyle L\left(P,f,\alpha\right)=0$. Now since $\displaystyle f\in\mathfrak{R}(\alpha)$ it follows that $\displaystyle \int_a^b f~d\alpha=\inf_{P}L\left(P,f,\alpha\right)=0$
Would this be a correct alternative solution to the above question? I did not using a lemma in my last one.

Answer: By definition we know that $\displaystyle f\in\mathfrak{R}(\alpha)$ iff for every $\displaystyle \varepsilon>0$ there exists a partition $\displaystyle P$ of $\displaystyle [a,b]$ such that $\displaystyle U\left(P,f,\alpha\right)-L\left(P,f,\alpha\right)<\varepsilon~~{\color{red} (*)}$. So if we can prove $\displaystyle \color{red}(*)$ for this particular function then we have proved integrability.

So now consider as I stated in my first solution that since for any subinterval $\displaystyle [x_{i-1},x_i]$ created by the partition $\displaystyle P$ there exists a $\displaystyle x\ne x_0$ an element of $\displaystyle [x_{i-1}x_i]$. This shows that $\displaystyle m_i=\inf_{[x_{i-1},x_i]}f=0$ which in turn implies that $\displaystyle L\left(P,f,\alpha\right)=0$ for any partition. Thus proving $\displaystyle \color{red}(*)$ reduces to proving $\displaystyle U\left(P,f,\alpha\right)<\varepsilon$.

To do this first define a paritition $\displaystyle P$ of $\displaystyle [a,b]$, we can see that $\displaystyle M_i=\sup_{[x_{i-1},x_i]}f=\left\{ \begin{array}{rcl} 1 & \mbox{if} & x_0\in[x_{i-1},x_i]\\ 0 & \mbox{if} & x\not\in[x_{i-1},x_i]\end{array}\right.$. So now let $\displaystyle [x_0-\delta,x_0+\delta]$ represent the interval centered at $\displaystyle x_0$, then it is apparent that $\displaystyle U\left(P,f,\alpha\right)=\sum_{i=1}^{n}M_i\cdot\De lta\alpha_i=\alpha\left( x_0+\delta\right)-\alpha\left( x_0-\delta\right)$. So now we can pick a parition that forces $\displaystyle \delta$ to be as small as we want, this in turn implies that we may pick a partition that makes $\displaystyle \alpha\left( x_0+\delta\right)-\alpha\left( x_0-\delta\right)$ as small as we want. This last fact is due to $\displaystyle \alpha$ being continuous at $\displaystyle x_0$. This proves that $\displaystyle f\in\mathfrak{R}(\alpha)$

So now for the value of the integral I stay with my previous response.

Im pretty dang sure that is correct...but with analysis I never know . Any criticism is very welcome!

9. Originally Posted by Mathstud28
So now for the value of the integral I stay with my previous response.
Sorry to beat a dead horse everyone, I just coming up with things about his basic problem. To compute the value of $\displaystyle \int_a^b f~d\alpha$ couldnt we consider this? Note, let us slightly generliaze this to $\displaystyle f:x\longmapsto\left\{\begin{array}{rcl} \xi & \mbox{if} & x=x_0\\ 0 & \mbox{if} & x\ne x_0\end{array}\right.$. It is clear that $\displaystyle f\in\mathbb{R}(\alpha)$ following the above arguments.

So now to find the value first consider that since $\displaystyle f\in\mathfrak{R}(\alpha)$ that for any interval containing $\displaystyle x_0$ that $\displaystyle \int_a^b f~d\alpha=\int_{x_0-\delta}^{x_0+\delta}f~d\alpha+\int_{a}^b f~d\alpha$ where $\displaystyle [c,d]$ is just "excess" interval not containing $\displaystyle x_0$. Now Consider that on any interval not containing $\displaystyle x_0$ $\displaystyle \int_a^b f~d\alpha=0$, so that $\displaystyle \int_a^b f~d\alpha=\int_{x_0-\delta}^{x_0+\delta}f~d\alpha$. Now note that repeated reasoning shows that then $\displaystyle \int_a^b f~d\alpha=\lim_{\delta\to0}\int_{x_0-\delta}^{x_0+\delta}f~d\alpha$. But now consider that since $\displaystyle f\geqslant0$ that $\displaystyle 0\leqslant\int_{x_0-\delta}^{x_0+\delta} f d\alpha\leqslant\xi\left[\alpha_(x_0+\delta)-\alpha(x_0-\delta)\right]$. So now then $\displaystyle 0\leqslant\int_a^b f~d\alpha=\lim_{\delta\to0}\int_{x_0-\delta}^{x_0+\delta}f~d\alpha\leqslant\lim_{\delta \to0}\xi\left[\alpha(x_0+\delta)-\alpha(x_0-\delta)\right]$ and since $\displaystyle \alpha$ is continuous at $\displaystyle x_0$ this implies that $\displaystyle 0\leqslant\int_a^b f~d\alpha\leqslant0\implies\int_a^b f~d\alpha=0$

Just wanted to share.