I have the following when I apply the dominated convergence theorem :

Let's say I want to find : $\lim_{n \to +\infty}\int_0^1 t^n \mathrm{d}t$ using the DCT (I know we can just integrate but it's an example to show where I have a problem).

So we defined the set of function : $f_n : [0,1] \to \mathbb{R} : t \mapsto t^n$. We have : $f_n \leq 1$ and $f_n$ converges to the piewise continuous function : $g = 0$ on $[0,1[$ and $g(1) = 1$.

Now the problem is that when I interchange limit and integration :

$$\lim_{n \to \infty} \int_0^1 f_n = \int_0^1 \lim_{n \to \infty} f_n = \int_0^1 g$$

Then clearly $\int_0^1 g(x) \mathrm{d}x = 0$ but how can I show this rigourously ? Since $g(1) = 1$ I don't know how to prove it's equal to $0$. Maybe I cant take an $\epsilon$ and say : $\int_0^1 g \leq \int_0^{1-\epsilon} 0 + \int_{1-\epsilon}^1 1 $

But it seems a lot of effort to prove an obvious fact.

Thank you !

N.B : All these integrals are Riemann not Lesbegue

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    $\begingroup$ These are Lebesgue integrals, right? It is a basic fact in Lebesgue integration theory that if $g=0$ almost everywhere then $\int g=0$. $\endgroup$ – Lorenzo Quarisa Mar 3 '19 at 13:12
  • $\begingroup$ @LorenzoQuarisa No i don't know about Lesbegue integrals. These are Riemann integrals. $\endgroup$ – dzhqjk Mar 3 '19 at 13:30
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    $\begingroup$ Your $\epsilon$ approach will work fine. Why do you say that it's a lot of work? $\endgroup$ – Michael Burr Mar 3 '19 at 13:34
  • $\begingroup$ @MichaelBurr Thank you. I was wondering if there is something simpler like for example defining the $f_n$ only on $[0,1[$ and not on $[0,1]$ ? Like that we get rid of the problem at $1$, yet I don't know if this is correct... $\endgroup$ – dzhqjk Mar 3 '19 at 13:36

First, you have to understand that the Lebesgue integral is a generalization of the Riemann integral. What we mean by that is that if a function is Riemann integrable, it is Lebesgue integrable and the two notions of integration are he same, ie : $\int_a^b f(x)dx=\int_{[a,b]}f dm$ where $m$ is the Lebesgue measure. A powerful theorem for this equivalence is the following : $f$ is Riemann integrable iff its set of discontinuities has Lebesgue measure 0. Now in your example, it is easy to see that you only have 1 point where the function is discontinuous and so the two notions agree. Suppose now you dont know that this theorem exists and that you only care about Riemann integrals. The important thing you have to remember is that integration never depends on the value of a function at a point ! Namely, you can change countably many points of your function (for example, let the function be equal to $\pi$ at all rational points !) and the resulting value of the integral will be the same. Your idea is good ! Simply take an $\epsilon$ neighborhood around 1, integrate as you wanted to do and let $\epsilon\to 0$. The intuition behind that is that the area under a single point is always as small as you want it to be ! I hope I clarified your uncertainties, if you have any questions leave them in the comment.

PS : The power of the Lebesgue Integration theory lives in its powerful convergence theorems (monotone convergence and dominated convergence). They not only yield results that were previously unobtainable with the Riemann theory of integration but also reduce the labor in proving classical results (such as a function being Riemann integrable).


To answer your question: It's trivial to verify from the definition of the integral that $\int_0^1 f(t)\,dt=0$.

But it's very curious that you talk about DCT in the context of Riemann integrals! Because the statement usually known as "DCT" is false for the Riemann integral - hence one wonders exactly what theorem you're referring to.

A weak version of DCT for integrals on $[0,1]$:

DCT. Suppose $f_n$ and $g$ are integrable on $[0,1]$, $|f_n|\le g$, and $f(t)=\lim_nf_n(t)$ exists for every $t\in[0,1]$. Then $f$ is integrable on $[0,1]$ and $\int_0^1 f=\lim_n\int_0^1f_n$.

Counterexample for the Riemann integral: In general if $A$ is a set define the function $\chi_A$ by $$\chi_A(t)=\begin{cases}1,&(t\in A), \\0,&(t\notin A).\end{cases}$$

Say $E=\{r_1,r_2,\dots\}$ is a countable dense subset of $[0,1]$. Let $E_n=\{r_1,\dots,r_n\}$. Let $f_n=\chi_{E_n}$, $g=1$. Then $$\lim_nf_n(t)=f(t)=\chi_E(t)$$for every $t$, but $f$ is not Riemann integrable (because for example it is discontinuous everywhere).

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    $\begingroup$ I think the DCT argument can easily be repaired for Riemann-integrals by assuming that the limit function must be integrable (as part of the hypothesis). $\endgroup$ – user370967 Mar 3 '19 at 16:15
  • $\begingroup$ That certainly gives a true theorem. Can you give an "easy" proof (without using the Lebesgue integral or similar cheating)? The question remains exactly what the OP means by "DCT", since the hypothesis that $f$ be integrable is not part of the theorem usually called DCT. (The fact that integrability is a conclusion in DCT is a large part of what makes iit useful.) $\endgroup$ – David C. Ullrich Mar 3 '19 at 16:22
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    $\begingroup$ I certainly can't give an easy proof without the theory of Lebesgue integral, but maybe the OP is given the statement I refer to (and which he calls DCT) without a proof. Seems the most plausible thing for me. $\endgroup$ – user370967 Mar 3 '19 at 16:33
  • $\begingroup$ Someone should explain the downvote... $\endgroup$ – David C. Ullrich Mar 3 '19 at 16:57
  • $\begingroup$ Here is my version of DCT (the one we've seen in class) for Riemann integrales : if $(f_n)_{n \in \mathbb{N}}$ is a sequence of piecewise continuous functions which converge to a continuous function $f$. Then if there is an integrable function $g$ such that : $\forall n, \forall x \in I, \mid f_n(x) \mid \leq g(x)$ then we can interchange limit and integral. We didn't see the proof in class since it's to complicated for our level. $\endgroup$ – dzhqjk Mar 3 '19 at 17:02

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