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Prove or disprove. Let $f_n(x): [0,1] \rightarrow \mathbb{R}$ be continuous and for every $x \in [0,1]$ suppose that we have

$$\lim_{n \rightarrow \infty} f_n(x)=0.$$

Then prove

$$\lim_{n \rightarrow \infty} \int_0^1 f_n(x)=0.$$

To prove this, I wanna say that since for each $n$, $f_n$ is continuous over a compact subset, then each $f_n$ is uniformly continuous. Since we are uniformly continuous can't I just push the limit inside and I get $0$? So its really more an argument of whether or not one can swap limit for integral.

homosapien
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    Counterexamples: https://math.stackexchange.com/q/1643629/42969, https://math.stackexchange.com/q/771957/42969, https://math.stackexchange.com/q/1575398/42969 – Martin R Dec 07 '21 at 19:42

1 Answers1

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You cannot prove it, since it is false. Just take $$f_n(x)=\begin{cases}n^2x&\text{ if }x\leqslant\frac1{2n}\\n^2\left(\frac1n-x\right)&\text{ if }x\in\left[\frac1{2n},\frac1n\right]\\0&\text{ otherwise.}\end{cases}$$Then, for each $x\in[0,1]$, $\lim_{n\to\infty}f_n(x)=0$. But$$(\forall n\in\Bbb N):\int_0^1f_n(x)\,\mathrm dx=\frac14.$$

José Carlos Santos
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  • woah, I wouldn't have come up with that. outta curiosity is this something that one see's typically as a trick ? I could never come up with that map. – homosapien Dec 07 '21 at 19:42
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    Various counterexamples have already been given on this site. – Martin R Dec 07 '21 at 19:44
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    Examples such as this one appear quit often. The graphs of the $f_n$'s are triangles with a base that gets narrower and a height the gets higher. And this is done in such a way that the are of such a triangle is $\frac14$. – José Carlos Santos Dec 07 '21 at 19:44
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    @HossienSahebjame There's nothing too special about this map AFAIK. Basically, you need it to $f_n(x)$ to have some kind of spike that gets taller and skinnier and moves closer and closer to the boundary as $n \rightarrow \infty$. The idea is that the area under the curve stays the same but is pushed away to the side so that the pointwise limit is still 0. It helps to draw a picture. – Jair Taylor Dec 07 '21 at 19:50
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    As a motivation for the counterexample you might want to know about the theorem of Arzelà and Osgood from the 1880s. They proved that if the sequence of continuous functions here is uniformly bounded and converging pointwise to zero then $\lim_{n\to\infty}\int_0^1 f_n(t),dt=0$. Knowing this you realize that your counterexample has to be unbounded. It is not that difficult then to construct one. – B. S. Thomson Dec 07 '21 at 20:14