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I found this question in my schools's Riemann's past question.

Let $\{f_n\}$ be a sequence of non-negative function which converges to an integrable function $f$ and supposing $ f_{n}\leq f(x)$ for each $n$. It is known that \begin{align} \int f(x)dx\leq \liminf\int f_n(x) dx\end{align} Then, show that \begin{align} \lim\limits_{n\to\infty}\int f_n(x)dx=\int \lim\limits_{n\to\infty}f_n(x)dx=\int f(x)dx\end{align}

QUESTIONS

  1. I believe this question is based on measure theory. Should this question appear at all in Riemann integration?

  2. Can we do this proof by Riemann integration? If no, then I'll need help but below is my trial.

MY TRIAL

Since $f_{n}\leq f(x)$ for each $n$, then \begin{align} \int \liminf f_n(x)dx \leq\int f(x)dx\end{align} I cannot interchange limit and integral since I am not sure if $f_n$ converges to $f$ uniformly. SO, I'm stuck here. Please, can someone help me? Mind you I have not been introduced to Measure theory yet but I can learn the steps if someone puts me through. That's if there's no other way but Measure theory approach. Thanks!

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    $\begingroup$ The first inequality is known as Fatou's lemma. It holds under even a weaker assumption that $f(x)=\liminf_{n\to\infty}f_n(x)$. Hence the conclusion cannot always hold, just by assuming something which is always true. $\endgroup$ – uniquesolution Sep 13 '18 at 7:11
  • $\begingroup$ uniquesolution: Hmm...? $\endgroup$ – Omojola Micheal Sep 13 '18 at 7:12
  • $\begingroup$ What is your question, Mike? $\endgroup$ – uniquesolution Sep 13 '18 at 7:12
  • $\begingroup$ @uniquesolution: I want to show that \begin{align} \lim\limits_{n\to\infty}\int f_n(x)dx=\int \lim\limits_{n\to\infty}f_n(x)dx=\int f(x)dx\end{align} $\endgroup$ – Omojola Micheal Sep 13 '18 at 7:13
  • $\begingroup$ @uniqueresolution: The title of the question is not optimal. What the OP is actually assuming is that $0\leq f_n \leq f$, which is quite a strong assumption. $\endgroup$ – PhoemueX Sep 13 '18 at 7:14
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What you are trying to prove is not true. Take $f_n(x)$ to be a function defined on $[0,\infty)$ which is zero everywhere except on the interval $[n-1/n,n+1/n]$, where its graph is a triangle whose vertices are at the points $(n-1/n,0),(n+1/n,0),(n,n)$. Then $f_n(x)$ converges pointwise to the zero function, but for every $n$, the integral of $f_n(x)$ is exactly $1$.

The formulation of the question seems to assume that the liminf inequality provides some information that may assist in proving the desired result, but that inequality does not provide any new information. In fact, it is true even under milder conditions - $f_n(x)$ need not converge pointwise to $f(x)$, and it is enough that $f(x)=\liminf f_n(x)$. In that setting it is known as Fatou's lemma. In other words, if one could prove the desired assertion, then it would follow that whenever $f_n(x)$ converges pointwise to $f(x)$, the integrals converge too, which - as is well known and shown by the above example -- is not true.

Addendum If you change the assumptions and assume that $f_n(x)\leq f(x)$, and that $f(x)$ is integrable, then the desired convergence of integrals is an immediate result of Lebesgue's dominated convergence theorem.

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  • $\begingroup$ I think I added a bit of my opinion. It should be $f_n(x)\leq f(x)$ for each $n.$ I'll edit it! $\endgroup$ – Omojola Micheal Sep 13 '18 at 7:23
  • $\begingroup$ I've corrected it! $\endgroup$ – Omojola Micheal Sep 13 '18 at 7:24
  • $\begingroup$ Ok, so in that case the answer is affirmative, as you can see in the latest addition. In any case, you should get rid of the confusing pseudo-assumption concerning the liminf. $\endgroup$ – uniquesolution Sep 13 '18 at 7:31
  • $\begingroup$ Does that mean that PhoemueX's hint would work? $\endgroup$ – Omojola Micheal Sep 13 '18 at 7:33
  • $\begingroup$ It seems that if you add enough assumptions you don't need to work at all. His hint was wrong when given, anyway. $\endgroup$ – uniquesolution Sep 13 '18 at 7:34

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