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

In summary: So if I may ask, what's your goal here? Are you looking to learn about some specific topic, or is this just a general curiosity?In summary, the conversation revolved around proving the Riemann-integrability of a function using uniform convergence and a proof attempt was discussed. The main concern was whether the proof was correct and how to expand it to show that the entire Riemann sum converges to the integral of the function. It was suggested to start with a simpler question and prove that the sequence of integrals converges to a number. A proof was presented using the definitions of Riemann and Darboux integrability and the fact that they are equivalent. The conversation ended with a discussion about the assumption of the
  • #1
Eclair_de_XII
1,083
91
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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  • #2
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
 
  • #3
Office_Shredder said:
The definition of your partition assumes ∫f exists!
Oh. I must have forgotten to subscript that.
 
  • #4
Office_Shredder said:
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.
 
  • #5
Using the fact that Riemann and Darboux integrability are equivalent, and that a function [itex]f : [a,b] \to \mathbb{R}[/itex] is Darboux integrable if and only if for every [itex]\epsilon > 0[/itex] there exists a partition [itex]D[/itex] of [itex][a,b][/itex] such that [itex]U(f,D) - L(f,D) < \epsilon[/itex] where [itex]U(f,D)[/itex] and [itex]L(f,D)[/itex] are the upper and lower sums of [itex]f[/itex] with respect to [itex]D[/itex]:

Let [itex]\epsilon > 0[/itex]. By uniform convergence of [itex]f_n \to f[/itex] on [itex][a,b][/itex] we have that for [itex]n[/itex] sufficiently large [tex]
\sup|f_n - f| < \frac{\epsilon}{3(b-a)}.[/tex] Since [itex]f_n[/itex] is integrable, there exists a partition [itex]D[/itex] of [itex][a,b][/itex] such that [itex]U(f_n,D) - L(f_n,D) < \frac13 \epsilon[/itex]. Hence [tex]
\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}[/tex] and [itex]f[/itex] is integrable.

To show that [itex]\int_a^b f_n\,dx \to \int_a^b f\,dx[/itex], let [itex]\epsilon > 0[/itex] and [itex]n[/itex] sufficiently large that [itex]\sup |f_n - f| < \epsilon/(b-a)[/itex]. Then [tex]
\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.[/tex]
 
  • #6
Eclair_de_XII said:
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.
 
  • #7
Office_Shredder said:
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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  • #8
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.
 

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

1. What does it mean for a sequence of functions to converge?

Convergence of a sequence of functions means that as the index of the sequence increases, the functions get closer and closer to a single limiting function. In other words, the values of the functions approach a fixed value as the sequence progresses.

2. How is the limit of a sequence of functions defined?

The limit of a sequence of functions is defined as the function that the sequence converges to as the index of the sequence approaches infinity. This can be thought of as the "end behavior" of the sequence of functions.

3. What is the importance of convergence in mathematics?

Convergence is important in mathematics because it allows us to study infinite sequences of functions and understand their behavior. It also plays a crucial role in many areas of mathematics, such as calculus, analysis, and differential equations.

4. Can a sequence of functions converge to more than one limit?

No, a sequence of functions can only converge to one limit. If a sequence of functions converges to more than one limit, it is not considered a convergent sequence.

5. How can we determine if a sequence of functions converges?

There are various methods for determining if a sequence of functions converges, such as the squeeze theorem, the comparison test, and the ratio test. These tests involve analyzing the behavior of the functions and their limiting values as the index of the sequence increases.

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