# Does $\displaystyle\lim_{n\to \infty} \int^{\infty}_{-\infty} f_n(x) dx\, = \int^{\infty}_{-\infty} f(x) dx\,$?

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

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

#### Solutions Collecting From Web of "Does $\displaystyle\lim_{n\to \infty} \int^{\infty}_{-\infty} f_n(x) dx\, = \int^{\infty}_{-\infty} f(x) dx\,$?"

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.

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