I have been stack with something which is perhaps just in front of my eyes but I just cannot get it. I have a sequence $(f_n)$ in $L^2(\mathbb R)$ which converges to $f$ in the $L^2$-topology. If I assume that $\int_{\mathbb R}f_n(x)dx=0$ for all $n$, then does it follow that $\int_{\mathbb R}f(x)dx=0$? Sorry if this too obvious.
Thanks for answer.
Math
On
Counterexample: Let $$f(x) = \frac{\operatorname{sgn}x}{1+|x|}.$$ Notice that $f$ is odd, and $f\in L^2\setminus L^1$. Then define $$f_n = \mathbb 1_{[-n,n]} f \in L^1\cup L^2.$$ and note that $f_n \to f$ in $L^2$. Finally, $\int f_n = 0$ for every $n$, but $|f|$ is not integrable, so $\int f$ does not exist.
Response to comment: If $xf_n$ converges in $L^2$, then it converges to $xf$. This is because $xf_n$ clearly converges to $xf$ in $L^2_\text{loc}$, say, which is a space with unique limits. Now define $\langle x\rangle^2 := 1+|x|^2$. Then simply by Holder's inequality,
$$\int |f| = \int\langle x\rangle|f| \frac{1}{\langle x\rangle}dx ≤ \left( \int \langle x\rangle^2|f|^2 dx\int \frac1{\langle x\rangle^2} dx\right)^{1/2} < \infty $$
so $f\in L^1$, and similarly
$$\int |f_n - f| \leq \left(\int \langle x\rangle^2 |f_n - f|^2 dx\int \frac1{\langle x\rangle^2}dx\right)^{1/2} \to 0$$
Hence, $$ \left|\int f_n - \int f\right| \leq \int |f_n - f| \to 0 $$ and $\int f = 0$.
Counterexample: suppose $f_n = \chi_{[0, 1]} - \frac{1}{n} \chi_{[1, n+1]}$. Then $f_n \to \chi_{[0, 1]}$ in $L^2(\mathbb{R})$. However, $\int_{\mathbb{R}} f_n(x)\,dx = 0$ for each $n$ while $\int_{\mathbb{R}} \chi_{[0,1]}(x)\,dx = 1$.