Is the set $\{f: \int_{\mathbb{R}} f \leq h\}$ closed in $L^2(\mathbb{R})$?

Question

If the sequence $\{f_n\}$ in $L^2(\mathbb{R})$ converges to $f$ in $L^2(\mathbb{R})$ and $\int_{\mathbb{R}}f_n(t)dt\leq h$ for some $h>0$ and all $n\in \mathbb{N}$; does $f$ satisfy $\int_{\mathbb{R}}f(t)dt\leq h$?

Answer 1

Not necessarily. Try $f_n=2\,\mathbf 1_{(0,1)}-\frac1n\mathbf 1_{(1,n+1)}$, and $f=2\,\mathbf 1_{(0,1)}$. Then the integral of $(f_n-f)^2$ is $\frac1n$ hence $f_n\to f$ in $L^2$, but the integral of $f_n$ is $1$ and the integral of $f$ is $2$.