Let $f \in L^1(\mathbb{R})$ and $\|f\|_1=1$, show that for $\delta>0$ $\lim_{n \to \infty}\int_{|x|\geq \delta}nf(nx)dx=0$

Question

Let $f \in L^1(\mathbb{R})$ and $\|f\|_1=1$, show that for $\delta>0$ $\lim_{n \to \infty}\int_{|x|\geq \delta}nf(nx)dx=0$

Also show that for $g$ bounded continuous, $\lim_{n \to \infty}\int_{\mathbb{R}}nf(nx)g(x)dx=g(0)$

I did it using $u-$sub enforcing $u=nx$ we get that $\int_{|x|\geq \delta}nf(nx)dx=\int_{n\delta}^{\infty}f(u)du+\int_{-\infty}^{-n\delta}f(u)du$ which goes to $0$ by DCT. Similarily we do the second part. We know by previous part that $\lim_{n \to \infty}\int_{\mathbb{R}}nf(nx)g(x)dx=\lim_{n \to \infty}\int_{-\delta}^{\delta}nf(nx)g(x)$ you do the u-sub, DCT + continuity give you the result.

I think my solution is valid but I have not proved the fact that $u$-sub is valid for infinite bounds (only for finite), and so I was wondering whether there is a non $u$-sub solution.

Answer 1

The substitution $x=u/n$ is valid on any integral over any measurable set, bounded or not.

There's an easier proof for the second limit:

$$\int_{\mathbb R}nf(nx)g(x)\,dx = \int_{\mathbb R}f(u)g(u/n)\,du.$$

As $n\to \infty,$ the integrands on the right $\to f(u)g(0)$ pointwise, by the continuity of $g$ at $0$ (you only need continuity at this one point). Since $|f(u)g(u/n)| \le |f(u)|\|g\|_\infty,$ the DCT gives the desired answer.