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.

Answer: Using the following lemma:

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.

Thanks in advance