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Let $f_n(x)$ and $f(x)$ be continuous functions on $[0, 1]$ such that $\lim_{n\to\infty} f_n(x) = f(x)$ for all $x \in [0, 1]$. Answer each of the following questions. If your answer is “yes”, then provide an explanation. If your answer is “no”, then give a counterexample.

(a) Can we conclude that $\lim_{n\to\infty}\int_{0}^{1}f_n(x)\,dx = \int_{0}^{1}f(x)\,dx$?

(b) If in addition we assume $|{f_n(x)}|\leq 2017$ for all n and for all $x \in [0, 1]$, can we conclude that $\lim_{n\to\infty}\int_{0}^{1}f_n(x)\,dx = \int_{0}^{1}f(x)\,dx$?

My attempt:

I could not think of any counterexample for part (a).

For part (b), I showed that ${f_n}$ is equicontinuous which along with uniform boundedness gives me that ${f_n}$ has a uniformly convergent subsequence. (Arzela Ascoli). Hence $\lim_{n\to\infty}\int_{0}^{1}f_n(x)\,dx= \lim_{k\to\infty}\int_{0}^{1}f_{n_k}(x)\,dx = \int_{0}^{1}f(x)\,dx$. Is that right?

Edit: I have made a mistake while proving equicontinuity.

I know that monotonicity + continuity + pointwise convergence on [0,1] ensures uniform convergence. Is it true that uniform boundedness + continuity + pointwise convergence ensures uniform convergence? How do I show that?

user557
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Xplorer
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    For part (a) see https://math.stackexchange.com/questions/2877869/how-to-construct-a-counterexample-to-lim-n-to-infty-int-01-f-nxdx-int-0/2877882#2877882 – Jacky Chong Aug 13 '18 at 04:25
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    What you've done for part $(b)$ is not quite right. You either need to show that all of $f_n$ converges to $f$ uniformly, or else that the limit $\lim_{n\to\infty}\int_{0}^{1}f_n(x),dx$ exists. If you don't know that the limit exists, it does not make sense to claim equality with the limit of a subsequence. – Fimpellizzeri Aug 13 '18 at 04:51
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    Actually it could be proved that if $f_n$ is equicontinuous and uniformly bounded then the convergence is uniform. – xbh Aug 13 '18 at 05:11
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    The hypothesis in part b) does not imply uniform convergence or equicontinuity. Example: $f_n(x)=nx$ for $0\leq x \leq \frac 1 n$, $f_n(x)=1-n(x-\frac 1 n)$ for $\frac 1 n\leq x \leq \frac 2 n$ . – Kavi Rama Murthy Aug 13 '18 at 05:42
  • @xbh Then Arzela Ascoli would guarantee uniform convergence of ${f_n}$, not just of a subsequence. – Xplorer Aug 13 '18 at 12:57
  • @Xplorer Here the function family is countable, and the interval is closed, you could try to prove this. Maybe the property of compact sets would help. – xbh Aug 13 '18 at 13:00
  • @xbh But Arzela Ascoli also holds for sequence of functions on compact sets. I probably did not understand you. – Xplorer Aug 13 '18 at 13:04
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    @Xplorer This question is a specific case for A-A theorem. The assumption is stronger than that of A-A: $f_n \to f$, which is not stated in A-A. – xbh Aug 13 '18 at 13:05
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    Also part b) is a simpler case for Arzela Dominated convergence theorem [whose proof is complicated]. As the comment by @K.R. Murthy shows, your method might need to be reconsidered. – xbh Aug 13 '18 at 13:17
  • @KaviRamaMurthy Thanks for your clarification! – xbh Aug 13 '18 at 13:18
  • By the way, how did you show that $f_n$ is equicontinuous? – Fimpellizzeri Aug 13 '18 at 15:07
  • @Fimpellizieri I realised I made a mistake while proving equicontinuity. I'll edit my question. – Xplorer Aug 13 '18 at 15:28
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    Yes, I think it's not true in general. Consider a tent sequence with fixed height, say $f_n(0) = 0$, then $f_n$ rises linearly to $1$ at $x=1/(2n)$, after which it goes back linearly to $0$ at $x=1/n$ and stays there. This sequence converges pointwise to the everywhere-zero function. The integrals do converge, but we don't have equicontinuity (inspect near $x=0$). This is also obvious a posteriori, since clearly this sequence does not converge uniformly to zero. – Fimpellizzeri Aug 13 '18 at 15:30
  • Since you know that monotonicty + continuity + pointwise convergence implies uniform convergence on compacts, you have a particular case of the monotone convergence theorem. You can use it to prove (a particular case of) Fatou's lemma, which is all you really need to prove (a particular case of) the dominated convergence theorem. Not sure how all of this fits into a qualifying exam question, though. – Fimpellizzeri Aug 14 '18 at 04:52
  • This question was exactly the reason why I asked the question @JackyChong mentions. For part $(b)$, you can take a look at what is called Arzela's Dominated Convergence Theorem (also see https://math.stackexchange.com/a/1779874/531299) – user557 Aug 14 '18 at 23:17
  • Oops, I haven't noticed that @xbh had already mentioned Arzela's DCT. (But you can find a link to its proof in the answer I referred to above.) Not sure if we're allowed to use this theorem though. – user557 Aug 14 '18 at 23:29

1 Answers1

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The question of switching limits is an important one, and with the Lebesgue integral one gets the powerful dominated convergence theorem. However, I presume you are using the Riemann integral.

For part a), I will give you a place to start. We denote the function $ \chi_S : \mathbb R \to \mathbb R $ via

\begin{equation*} \chi_S(x) = \begin{cases} 1 & x \in S \\ 0 & x \notin S \end{cases} \end{equation*}

For any subset $ S \subseteq \mathbb R $.

Let $ f_n = n \chi_{(0, \frac{1}{n})} $. Check that $ f_n(x) \to 0 $ for all $ x \in [0, 1] $, so $ f(x) = 0 $. Hint: do this first for $ x = 0 $, then for $ x \in (0, 1] $. It is easy to see that $ \int_{0}^{1} f_n(x) \,dx = 1 $ for all $ n $. Then $ f_n \to f $ pointwise, but $ \int_{0}^{1} f_n(x) \,dx \not\to 0 = \int_{0}^{1} f(x) \,dx $. These functions are not continuous, so I will leave it to you to modify these to get a continuous version of this counterexample. In fact, it is possible to find a counterexample where each function is smooth.

For part b), your proof is close, but you need more detail to fill it out. Specifically, you have shown that $ f_{n_k} $ converges uniformly to $ f $, but you have not shown this for $ f_n $. To do this, you must use the fact that $ f_n \to f $ pointwise, as this is false otherwise. Suppose then that $ f_n $ does not converge uniformly to $ f $. Then, there exists an $ \epsilon > 0 $ such that for some $ n_m \to \infty $, there exists a sequence $ x_{n_m} $ in $ [0, 1] $ such that $ |f_{n_m}(x_{n_m}) - f(x_{n_m})| \geq \epsilon $. We get this by negating the definition of uniform convergence. However, this tells us that no subsequence of $ f_{n_m} $ converges uniformly to $ f $, but Arzela Ascoli guarantees the existence of a uniformly convergent subsequence, and the pointwise convergence of $ f_n $ to $ f $ tells us that this uniformly convergent subsequence must converge to $ f $. This is a contradiction, so $ f_n \to f $ uniformly, so $ \int_0^1 f_n(x) \, dx \to \int_0^1 f(x) \, dx $.

paul blart math cop
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  • How can it be done without assuming equicontinuity? (since the sequence need not be equicontinuous). – Xplorer Aug 13 '18 at 15:55
  • Are you allowed to use Lebesgue integration? – Fimpellizzeri Aug 13 '18 at 17:23
  • No, we aren't. This is a qualifying exam question. – Xplorer Aug 13 '18 at 17:31
  • I believe equicontinuity may be necessary, as it is not true that if $f_n\tof$ pointwise and some subsequence converges uniformly to $f$ that $f_n\tof$ uniformly. For example, take some sequence $f_n\tof$ pointwise but not uniformly. Then let $g_{2n} = f_n$, $g_{2n+1} = f$. Then $g_n\tof$ pointwise, $g_{2n+1}\tof$ uniformly, but $g_n\not\tof$ uniformly. – paul blart math cop Aug 13 '18 at 19:24
  • @Xplorer I would think if it's a qualifying exam question it would be expected that hte integral is in the Lebesgue sense – operatorerror Aug 15 '18 at 02:29
  • @qbert If it's an analysis qual I would agree, but if it's a general qual then I would stick to the Riemann integral, as not all undergrad curriculums cover measure theory. – paul blart math cop Aug 15 '18 at 02:35
  • @qbert It's not in the Lebesgue sense. Lebesgue integrals and/or measure theory are not in the syllabus. – user557 Aug 15 '18 at 14:35