Let f_n denote an element in a sequence of functions that converges

[Eclair_de_XII]

TL;DR Summary

uniformly to $f$, where $f$ is some function. Suppose also that each $f_n$ is Riemann-integrable. Show that $f$ is Riemann-integrable also, and that the integral of $f$ is equal to that of $f_n$ whenever $n$ is sufficiently large.

Let $\epsilon>0$. Choose $N\in\mathbb{N}$ s.t. for each integer $n$ s.t. $n\geq N$,

$$|\sup\{|f-f_n|(x):x\in D\}|<\frac{\epsilon}{3}$$

where $D$ denotes the intersection of the domains of $f$ and $f_n$.

Choose a partition $P:=\{x_0,\ldots,x_m\}$ with $x_i<x_{i+1}$ where $0\leq i\leq m-1$ s.t. the following holds.

%%% EDIT: The function to be integrated has been corrected from $f$ to $f_n$.

$$|f_n(s_i)-f_n(t_i)|\Delta x_i<\frac{\epsilon}{3}$$
$$\Delta x_i\leq 1$$
$$|f_n(t_i)\Delta x_i -\int_{x_i}^{x_{i+1}} f_n|<\frac{\epsilon}{3}$$

where $s_i,t_i\in[x_i,x_{i+1}]$ and $\Delta x_i:=x_{i+1}-x_i$.

Hence,

$$\eqalign{%
\epsilon>&|f_n(s_i)-f(t_i)|\Delta x_i\cr
+&||f-f_n||+|f_n(s_i)\Delta x_i-\int_{x_i}^{x_{i+1}}f_n|\cr
\geq&|f_n(s_i)-f_n(t_i)|\Delta x_i\cr
+&|f(s_i)-f_n(s_i)|+|f_n(t_i)\Delta x_i - \int_{x_i}^{x_{i+1}}f_n|\cr
\geq&|(f_n(s_i)-f_n(t_i))\Delta x_i\cr
+&(f(s_i)+f_n(s_i))\Delta x_i\cr
+&(f_n(t_i)\Delta x_i-\int_{x_i}^{x_{i+1}} f_n)|\cr
=&|f(s_i)\Delta x_i-\int_{x_i}^{x_{i+1}}f_n|}$$

%%%

My main worry about this proof attempt is that it only shows that a rectangle with height equal to $f(s_i)$ and base $\Delta x_i$ has an area that is within $\epsilon$ units of the area under the graph of $f$ over the region $[x_i,x_{i+1}]$. But I'm not sure how to expand the proof, if it is correct, in order to show that the area of the entire Riemann sum converges to the integral of $f$.

Ideally, I would start by building the partition $P$ one point at a time so that the inequality above will hold. But the problem is that I wouldn't know beforehand how many partitions of $[x_0,x_m]$ would be needed. Maybe I could have each partition be of the same length sufficient in order to ensure convergence, then calculate it that way. But I'm a bit hesitant about implementing this idea; what if there is no uniform partition length that will ensure that the area of the rectangle converges to the area under $f$ over that partition?

 

[Office_Shredder]

The definition of your partition assumes $\int f$ exists!

The problem is also terribly stated, the integrals are probably never going to be exactly equal.

I think you should start with a simpler question. Prove the sequence $\int_a^b f_n(x) dx$ converges to a number

 

[Eclair_de_XII]

The definition of your partition assumes ∫f exists!

Oh. I must have forgotten to subscript that.

 

[Eclair_de_XII]

Prove the sequence $\int_a^b f_n(x) dx$ converges to a number

Choose $N$ big enough. Let $m$ be an integer greater than $N$. Then $\sup\{|f_m-f|(x)\}\rightarrow0$.Since the integral of a non-negative function is non-negative, and $m$ is large enough,

$$\int|f-f_m|\rightarrow0$$

Also,

$$|\int f - \int f_m| \leq \int|f-f_m|$$

so $\{\int f_i\}_{i\in\mathbb{N}}$ converges to $\int f$. By the triangle inequality, the marked sum in my opening post must converge to that value as well. Hence, $f$ is Riemann-integrable, I think.

 

[pasmith]

Using the fact that Riemann and Darboux integrability are equivalent, and that a function $f : [a,b] \to \mathbb{R}$ is Darboux integrable if and only if for every $\epsilon > 0$ there exists a partition $D$ of $[a,b]$ such that $U(f,D) - L(f,D) < \epsilon$ where $U(f,D)$ and $L(f,D)$ are the upper and lower sums of $f$ with respect to $D$:

Let $\epsilon > 0$. By uniform convergence of $f_n \to f$ on $[a,b]$ we have that for $n$ sufficiently large $$\sup|f_n - f| < \frac{\epsilon}{3(b-a)}.$$ Since $f_n$ is integrable, there exists a partition $D$ of $[a,b]$ such that $U(f_n,D) - L(f_n,D) < \frac13 \epsilon$. Hence $$\begin{split} U(f,D) - L(f,D) &< \left(U(f_n,D) + \tfrac13\epsilon\right) - \left(L(f_n,D) - \tfrac13\epsilon\right) \\ &= U(f_n,D) - L(f_n,D) + \tfrac23\epsilon \\ &< \tfrac13 \epsilon + \tfrac23 \epsilon \\ &= \epsilon\end{split}$$ and $f$ is integrable.

To show that $\int_a^b f_n\,dx \to \int_a^b f\,dx$, let $\epsilon > 0$ and $n$ sufficiently large that $\sup |f_n - f| < \epsilon/(b-a)$. Then $$\left|\int_a^b f_n(x)\,dx - \int_a^b f(x)\,dx\right| \leq \int_a^b |f_n(x) - f(x)|\,dx \leq (b-a)\sup |f_n - f| < \epsilon.$$

 

[Office_Shredder]

Choose $N$ big enough. Let $m$ be an integer greater than $N$. Then $\sup\{|f_m-f|(x)\}\rightarrow0$.Since the integral of a non-negative function is non-negative, and $m$ is large enough,

$$\int|f-f_m|\rightarrow0$$

Also,

$$|\int f - \int f_m| \leq \int|f-f_m|$$

so $\{\int f_i\}_{i\in\mathbb{N}}$ converges to $\int f$. By the triangle inequality, the marked sum in my opening post must converge to that value as well. Hence, $f$ is Riemann-integrable, I think.

Sorry I wasn't clear. Prove it without assuming that $\int f$ exists! This was me proposing a first step in the proof.

 

[Eclair_de_XII]

Prove it without assuming that $\int f$ exists!

When was it assumed that $\int f$ exists? All I deduced was that $\int |f_m-f|$ exists, because the index $m$ is sufficiently large, meaning that $|f_m-f|(x)\leq0$ [a fact that I now realize does not actually follow from the definition of uniform convergence], and because the function $|f_m-f|$ is non-negative. Then I used the inequality $|\int g|\leq|\int|g|$ for every integrable function $g$.

I'm not sure if I have the energy to formulate a proper/correct proof right now. And in any case, I don't very much see the point, seeing as someone has gone through the liberty of doing that already.

 

[Office_Shredder]

You don't know that $|f_n-f|$ is integrable, as one example $f_n=\frac{1}{n} \mathbb{1}_{\mathbb{Q}}$ uniformly converges to $f=0$, but $|f_n-f|$ is not integrated.

(Obviously the $f_n$ are not integrable, but we know a true counterexample doesn't exist).

The idea I was thinking of was that you can prove the sequence is Cauchy.

If you're happy with the outcome of the thread no need to worry more about my proposal.