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Consider a sequence of functions $f_n$ defined on $[0,1]$ such that $f_n=\frac{x^n}{1+x^n}$. Is it true that $\lim_{n\to \infty} \int_{0}^1 f_n(x)dx=\int_0^1 \lim_{n\to \infty} f_n(x)dx$?

My first attempt is to prove that $\{f_n\}$ uniformly converges to some $f$. But it is not. Is there any way to do this?

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  • $\begingroup$ Try evaluating each in terms of x and n. Then compare the two. If they're algebraically the same, you're done. $\endgroup$ – Jason Chen Mar 22 '17 at 6:16
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It seems to me exaggerated to use the DCT here, because this is a deep theorem and the example is completely elementary.

To be more precise, I would say that it is, of course, a perfect solution mathematically speaking, but perhaps not pedagogically speaking ...

Indeed, we have :

$$\forall x\in[0,1],\lim_{n\to\infty}f_n(x)=f(x)\textrm{, where }f(x)=\left\{\matrix{0&\textrm{if }0\le x<1\cr\frac12&\textrm{if }x=1}\right.$$

Hence :

$$\int_0^1\lim_{n\to\infty}f_n(x)\,dx=0$$

On the other hand :

$$\forall n\in\mathbb{N},\,0\le f_n(x)\le x^n$$

which leads to :

$$\forall n\in\mathbb{N},\,0\le\int_0^1f_n(x)\,dx\le\frac1{n+1}$$

and so :

$$\lim_{n\to\infty}\int_0^1f_n(x)\,dx=0$$

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Dominated convergence theorem gives this equality easily. Note for all $x \in [0,1], n \ge 1$, $|f_n(x)| \le 1$ which is integrable on $[0,1]$ and $f_n(x) \to 0$ as $n \to \infty$ for each $x \in (0,1)$.

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  • $\begingroup$ you mean $f_n$ converges uniformly to 0? $\endgroup$ – Kenneth.K Mar 22 '17 at 6:26
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    $\begingroup$ No. $(f_n)$ does not converge uniformly ! All $f_n$ are continuous. For $x \in [0,1)$ we have: $f_n(x) \to 0$ and for $x=1$ we have $f_n(1)=1/2 \to 1/2$. Hence the limit fuction $f(x)=\lim_{n \to \infty}f_n(x)$ is not continuous on $[0,1]$ $\endgroup$ – Fred Mar 22 '17 at 6:29

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