1. ## Analysis, integral

My book gave this as an excercise and I did it with a method similar to one given as an example, but I am curious of this alternate method will work. Forgive me if the proof is incoherent or just is wrong, I haven't slept in two days (long story).

Question: Let $\displaystyle f:x\longmapsto\left\{ \begin{array}{rcl} \frac{1}{2^n} & \mbox{if} & x=\frac{a}{2^n}~~a,n\in\mathbb{N}~2\not| a\\ 0 & \mbox{if} & \text{otherwise} \end{array}\right.$. Is $\displaystyle f\in\mathfrak{R}(\alpha)$ on $\displaystyle [0,1]$ and if so what is the value of $\displaystyle \int_0^1 f~d\alpha$

Notation: For my sake let $\displaystyle S=\left\{x:\frac{a}{2^n}~~a,n\in\mathbb{N}~2\not| a\right\}$

Answer: By definition we must show that $\displaystyle \forall \varepsilon>0$ there exists a partition $\displaystyle P$ of $\displaystyle [0,1]$ such that $\displaystyle U\left(P,f,\alpha\right)-L\left(P,f,\alpha\right)<\varepsilon~~{\color{red} (*)}$. Now it is clear that since on any interval $\displaystyle [x_{i-1},x_i]$ there exists values of $\displaystyle x$ such that $\displaystyle x\not\in S$, so consequently $\displaystyle m_i=\inf_{[x_{i-1},x_i]}~f=0$ which in turn implies that for any partiton $\displaystyle L\left(P,f,\alpha\right)=\sum_{i=1}^{n}m_i\cdot \Delta \alpha_i=0$. This realization enables to prove $\displaystyle \color{red}(*)$ by proving that $\displaystyle U\left(P,f,\alpha\right)<\varepsilon$.

To do this we must merely construct our partition carefully. Before we do this let me establish some "new" notation.

Define $\displaystyle \mathcal{O}_{\varphi}\left(x_i\right)$ as being a neighborhood of radius $\displaystyle \varphi$ around $\displaystyle x_i$

Now construct a partition of $\displaystyle [0,1]$ as follows: Let a partition $\displaystyle P_n=\left\{x_i\right\}~0\leqslant x\leqslant n$ of $\displaystyle [0,1]$ posses the following charcteristics $\displaystyle x_0=a$,$\displaystyle x_n=b$, and $\displaystyle x_i\not\in\mathbb{Q}~~1\leqslant i\leqslant n-1$. Now it is clear that we must have $\displaystyle 2\sum_{i} \mathcal{O}_{\varphi}\left(x_i\right)=1$.

So now it is obvious that we can make the $\displaystyle 2^n$ in $\displaystyle \frac{a}{2^n}$ as big as we want by making $\displaystyle \varphi$ as small as we want, so we can make $\displaystyle M_i=\sup_{[x_{i-1},x_i]}~f$ similarly as small as we want. Finally we can make $\displaystyle \sup_{i}\left(M_i\right)$ as small as we want.

Now let $\displaystyle \delta$ be a number such that $\displaystyle \varphi<\delta\implies \sup_{i}\left(M_i\right)\left(\alpha(b)-\alpha(a)\right)<\varepsilon$. Choose a partition of $\displaystyle [0,1]$ such that $\displaystyle \max\left\{\varphi\right\}<\delta$. Let $\displaystyle \sup_{i}\left(M_i\right)=M$, then

\displaystyle \begin{aligned}U\left(P,f,\alpha\right)&=\sum_{i=1 }^{n}M_i\cdot\Delta \alpha_i\\ &\leqslant M\sum_{i=1}^{n}\Delta\alpha_i\\ &=M\left(\alpha(b)-\alpha(a)\right)\\ &<\varepsilon\end{aligned}.

This proves integrability of $\displaystyle f$. Now to find the value just note that since it is integrable that $\displaystyle \int_0^1 f~d\alpha=\sup_{P}L\left(P,f,\alpha\right)=0$

Sorry if that is really far off.

2. Originally Posted by Mathstud28
My book gave this as an excercise and I did it with a method similar to one given as an example, but I am curious of this alternate method will work. Forgive me if the proof is incoherent or just is wrong, I haven't slept in two days (long story).

Question: Let $\displaystyle f:x\longmapsto\left\{ \begin{array}{rcl} \frac{1}{2^n} & \mbox{if} & x=\frac{a}{2^n}~~a,n\in\mathbb{N}~2\not| a\\ 0 & \mbox{if} & \text{otherwise} \end{array}\right.$. Is $\displaystyle f\in\mathfrak{R}(\alpha)$ on $\displaystyle [0,1]$ and if so what is the value of $\displaystyle \int_0^1 f~d\alpha$

Notation: For my sake let $\displaystyle S=\left\{x:\frac{a}{2^n}~~a,n\in\mathbb{N}~2\not| a\right\}$

Answer: By definition we must show that $\displaystyle \forall \varepsilon>0$ there exists a partition $\displaystyle P$ of $\displaystyle [0,1]$ such that $\displaystyle U\left(P,f,\alpha\right)-L\left(P,f,\alpha\right)<\varepsilon~~{\color{red} (*)}$. Now it is clear that since on any interval $\displaystyle [x_{i-1},x_i]$ there exists values of $\displaystyle x$ such that $\displaystyle x\not\in S$, so consequently $\displaystyle m_i=\inf_{[x_{i-1},x_i]}~f=0$ which in turn implies that for any partiton $\displaystyle L\left(P,f,\alpha\right)=\sum_{i=1}^{n}m_i\cdot \Delta \alpha_i=0$. This realization enables to prove $\displaystyle \color{red}(*)$ by proving that $\displaystyle U\left(P,f,\alpha\right)<\varepsilon$.

To do this we must merely construct our partition carefully. Before we do this let me establish some "new" notation.

Define $\displaystyle \mathcal{O}_{\varphi}\left(x_i\right)$ as being a neighborhood of radius $\displaystyle \varphi$ around $\displaystyle x_i$

Now construct a partition of $\displaystyle [0,1]$ as follows: Let a partition $\displaystyle P_n=\left\{x_i\right\}~0\leqslant x\leqslant n$ of $\displaystyle [0,1]$ posses the following charcteristics $\displaystyle x_0=a$,$\displaystyle x_n=b$, and $\displaystyle x_i\not\in\mathbb{Q}~~1\leqslant i\leqslant n-1$. Now it is clear that we must have $\displaystyle 2\sum_{i} \mathcal{O}_{\varphi}\left(x_i\right)=1$.

So now it is obvious that we can make the $\displaystyle 2^n$ in $\displaystyle \frac{a}{2^n}$ as big as we want by making $\displaystyle \varphi$ as small as we want, so we can make $\displaystyle M_i=\sup_{[x_{i-1},x_i]}~f$ similarly as small as we want. Finally we can make $\displaystyle \sup_{i}\left(M_i\right)$ as small as we want.

Now let $\displaystyle \delta$ be a number such that $\displaystyle \varphi<\delta\implies \sup_{i}\left(M_i\right)\left(\alpha(b)-\alpha(a)\right)<\varepsilon$. Choose a partition of $\displaystyle [0,1]$ such that $\displaystyle \max\left\{\varphi\right\}<\delta$. Let $\displaystyle \sup_{i}\left(M_i\right)=M$, then

\displaystyle \begin{aligned}U\left(P,f,\alpha\right)&=\sum_{i=1 }^{n}M_i\cdot\Delta \alpha_i\\ &\leqslant M\sum_{i=1}^{n}\Delta\alpha_i\\ &=M\left(\alpha(b)-\alpha(a)\right)\\ &<\varepsilon\end{aligned}.

This proves integrability of $\displaystyle f$. Now to find the value just note that since it is integrable that $\displaystyle \int_0^1 f~d\alpha=\sup_{P}L\left(P,f,\alpha\right)=0$

Sorry if that is really far off.
this proof is not correct, because for example since $\displaystyle \frac{1}{2} \in [x_i, x_{i+1}],$ for some i, we'll have $\displaystyle M_i=\frac{1}{2}.$ so we can't make all $\displaystyle M_j$ as small as we want. i think (i'm not sure right now) that we also need $\displaystyle \alpha$

to be continuous at least at elements of $\displaystyle S$ in order to make $\displaystyle \Delta \alpha_i$ as small as we want. you choose $\displaystyle n$ large enough so that $\displaystyle \frac{1}{2^n} < \epsilon.$ now let $\displaystyle x_i=\frac{i}{2^n}, \ 1 \leq i \leq 2^{n-1}.$ choose disjoint neighbourhoods

$\displaystyle U_i=[a_i,b_i]$ of $\displaystyle x_i$ such that $\displaystyle \sum_i \Delta \alpha_i < \epsilon.$ then $\displaystyle \sum_i (\sup f)\Delta \alpha_i < \frac{\epsilon}{2}.$ let $\displaystyle V_i=[b_{i-1}, a_i], \ 1 \leq i \leq 2^{n-1} + 1,$ with $\displaystyle b_0=0, \ a_{2^{n-1} + 1}=1.$ finally over $\displaystyle \bigcup_i V_i$ we have:

$\displaystyle \sum (\sup f) \Delta \alpha_i \leq 2^{-n} ( \alpha(b) - \alpha(a)) < \epsilon( \alpha(b)- \alpha(a)).$

3. Originally Posted by NonCommAlg
this proof is not correct, because for example since $\displaystyle \frac{1}{2} \in [x_i, x_{i+1}],$ for some i, we'll have $\displaystyle M_i=\frac{1}{2}.$ so we can't make all $\displaystyle M_j$ as small as we want. i think (i'm not sure right now) that we also need $\displaystyle \alpha$

to be continuous at least at elements of $\displaystyle S$ in order to make $\displaystyle \Delta \alpha_i$ as small as we want. you choose $\displaystyle n$ large enough so that $\displaystyle \frac{1}{2^n} < \epsilon.$ now let $\displaystyle x_i=\frac{i}{2^n}, \ 1 \leq i \leq 2^{n-1}.$ choose disjoint neighbourhoods

$\displaystyle U_i=[a_i,b_i]$ of $\displaystyle x_i$ such that $\displaystyle \sum_i \Delta \alpha_i < \epsilon.$ then $\displaystyle \sum_i (\sup f)\Delta \alpha_i < \frac{\epsilon}{2}.$ let $\displaystyle V_i=[b_{i-1}, a_i], \ 1 \leq i \leq 2^{n-1} + 1,$ with $\displaystyle b_0=0, \ a_{2^{n-1} + 1}=1.$ finally over $\displaystyle \bigcup_i V_i$ we have:

$\displaystyle \sum (\sup f) \Delta \alpha_i \leq 2^{-n} ( \alpha(b) - \alpha(a)) < \epsilon( \alpha(b)- \alpha(a)).$
Thanks NonCommAlg...in my state of coffee induced semi-catatonia I was thinking of $\displaystyle \sup$ as being the biggest $\displaystyle 2^n$

Can you also use this proof assuming that $\displaystyle \alpha$ is continuous?

Consider a partition $\displaystyle P_n$ of $\displaystyle [0,1]$ containing $\displaystyle n$ elements. Since $\displaystyle \alpha$ is continuous define $\displaystyle \Delta \alpha_i=\frac{\alpha(b)-\alpha(a)}{n^2}$. Now as you stated $\displaystyle M_i\leqslant\frac{1}{2}$. So

\displaystyle \begin{aligned}U\left(P_n,f,\alpha\right)&=\sum_{i =1}^{n}M_i\cdot\Delta\alpha_i\\ &\leqslant \frac{1}{2}\sum_{i=1}^{n}\Delta\alpha_i\\ &=\frac{\alpha(b)-\alpha(a)}{2n^2}\sum_{i}^{n}1\\ &=\frac{\alpha(b)-\alpha(a)}{2n}<\varepsilon\end{aligned}

for sufficiently large $\displaystyle n$. And since $\displaystyle U\left(P_n,f,\alpha\right)-L\left(P_n,f,\alpha\right)=U\left(P_n,f,\alpha\rig ht)<\varepsilon$ we have that $\displaystyle f\in\mathfrak{R}(\alpha)$. My answer still stands that $\displaystyle L\left(P_n,f,\alpha\right)=0\implies\int_0^1 f~d\alpha=\sup_{P_n}L\left(P_n,f,\alpha\right)=0$

Am I assuming something obviously not true in my above proof?

EDIT: Yes, I see what is wrong...it would be $\displaystyle \sum_{i=1}^{n^2}$...I understand your proof so Ill just let this be lesson learned unless I think of something better.

If it makes up for it, I think I realized that we can strengthen one of the theorems in my book...I think we can show that if $\displaystyle \sum f$ is bounded and $\displaystyle \alpha$ is continuous then $\displaystyle f\in\mathfrak{R}(\alpha)$

4. Hello everyone and thank you for reading this. On my way home from school I think I came up with another way to prove this. If any of the more knowledgable members here could take a look I would greatly appreciate it.

Question: Let $\displaystyle f:x\longmapsto\left\{ \begin{array}{rcl} \frac{1}{2^n} & \mbox{if} & x=\frac{a}{2^n}~~a,n\in\mathbb{N}~2\not| a\\ 0 & \mbox{if} & \text{otherwise} \end{array}\right.$. Is $\displaystyle f\in\mathfrak{R}(\alpha)$ on $\displaystyle [0,1]$ and if so what is the value of $\displaystyle \int_0^1 f~d\alpha$

Notation: For my sake let $\displaystyle S=\left\{x:\frac{a}{2^n}~~a,n\in\mathbb{N}~2\not| a\right\}$

Addendum: Let $\displaystyle \alpha$ be continuous on $\displaystyle [0,1]$
Just as before we want to show that for every $\displaystyle \varepsilon>0$ there exists a partition $\displaystyle P$ of $\displaystyle [0,1]$ such that $\displaystyle U\left(P,f,\alpha\right)-L\left(P,f,\alpha\right)<\varepsilon\quad{\color{r ed}(*)}$.

Now note that since given any interval of $\displaystyle [0,1]$ there must exist $\displaystyle x\not\in S$ in that interval, which implies that for any interval $\displaystyle m_i=\inf_{[x_{i-1},x_i]}f=0$ which in turn implies that $\displaystyle L\left(P,f,\alpha\right)=\sum_{i=1}^{n}m_i\cdot\De lta\alpha_i=0$. Due to this we can see that $\displaystyle \color{red}(*)$ is equivalent to showing that $\displaystyle U\left(P,f,\alpha\right)<\varepsilon$

We prove this as follows: For any partition $\displaystyle P_{n+1}$ of $\displaystyle [0,1]$ containing $\displaystyle n+1]$ elements define $\displaystyle \Delta \alpha_i=\frac{\alpha(b)-\alpha(a)}{n}$, this can be done by the continuity of $\displaystyle \alpha$. Now we can see that the partition $\displaystyle P_{n+1}$ divides $\displaystyle [0,1]$ into $\displaystyle n$ subintervals. Now it can be seen that

\displaystyle \begin{aligned}U\left(P_{n+1},f,\alpha\right)&=\su m_{i=1}^{n}M_i\cdot\Delta \alpha_i\\ &=\frac{\alpha(b)-\alpha(a)}{n}\sum_{i=1}^{n}M_i\\ &\leqslant\frac{\alpha(b)-\alpha(a)}{n}\sum_{i=1}^{n}\frac{1}{2^n}\\ &\leqslant\frac{2\left(\alpha(b)-\alpha(a)\right)}{n}<\varepsilon\end{aligned}

For sufficiently large enough $\displaystyle n$

Is this correct? Or am I misinterpreting the function?