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

• B
• Eclair_de_XII

#### 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?

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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

The definition of your partition assumes ∫f exists!
Oh. I must have forgotten to subscript that.

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.

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 $$\left|\int_a^b f_n(x)\,dx - \int_a^b f(x)\,dx\right| \leq \int_a^b |f_n(x) - f(x)|\,dx \leq (b-a)\sup |f_n - f| < \epsilon.$$

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.

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.

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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.