If $\displaystyle\{f_n(x)\}$ is sequence of continuous function on $\mathbb{R}$ converging uniformly to $f(x)$, then does $\displaystyle\lim_{n\to \infty} \int^{\infty}_{-\infty} f_n(x) dx\, = \int^{\infty}_{-\infty} f(x) dx\,$ hold?

I know it holds if we integrate over an interval of the form $[a,b]$ but not sure about improper integral!

  • $\begingroup$ $f_n$ uniformly converges on $\mathbb{R}$ $\endgroup$ – Bhauryal Jun 22 '14 at 17:26

Take $f_n$ a continuous function whose values on $[-n,n]$ is $1/n$, $f_n(x)=0$ if $\left|x\right|\gt n+1$ and $0\leqslant f_n(x)\leqslant 1/n$. Then $\sup_{x\in\mathbf R}\left|f_n(x)\right|=1/n$ from which we deduce that $f_n\to f\equiv 0$ uniformly on the real line. But since $f_n$ is non-negative, $\int_{\mathbf R}f_n(x)\mathrm dx\geqslant \int_0^nf_n(x)\mathrm dx=1$.

In other word, a uniform control is not enough if we integrate over a set of infinite measure.

  • $\begingroup$ Of course, if $x\in I$ and $f(x,y)$ is continuous for all $x\in I$ and $y\in \mathbb{R}$, and if the integral $F(x)=\int_{\mathbb{R}}f(x,y)\,dy$ converges uniformly for $x\in I$, then $F(x)$ is continuous on $I$. How is $f_n(x)=1/n$ for $|x|\le n$ and $f_n(x)=0$ for $x>n$ continuous at $x=n$? $\endgroup$ – Mark Viola Mar 10 '17 at 4:14

$$f(x)= \frac{1}{n} \chi_{[n,2n]}$$ will also do.

  • 1
    $\begingroup$ It works, but it's not continuous as you have required so you would need to modify $f_n(x)$ close to $ x= n$ and $x=2n$ to make it continuous. btw this is basically the same function as in the other answer $f_n\sim \frac{1}{n}\chi_{[-n,n]}$. $\endgroup$ – Winther Jun 9 '16 at 21:11
  • $\begingroup$ see this terrytao.wordpress.com/2010/10/02/… $\endgroup$ – reuns Jun 9 '16 at 23:56

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