This should be a relatively easy question, but I can't seem to figure it out. I want to show that $S = \{f \in L^1(\mathbb{R}) \mid \int_{\mathbb{R}}f dm = 0\}$ is closed in $L^1(\mathbb{R})$.
As $L^1(\mathbb{R})$ is a normed space, it is enough to show that if $f_n \to f$ in $L^1(\mathbb{R})$, then $f \in S$. I was hoping to use on of the convergence theorems, but this sequence isn't necessarily monotone or dominated. As $f_n \to f$ in $L^1(\mathbb{R})$ there is a subsequence $f_{n_k}$ such that $f_{n_k} \to f$ a.e. In particular, $f_{n_k}^+ \to f^+$ a.e. and $f_{n_k}^- \to f^-$ a.e. so by Fatou's Lemma we have $\int_{\mathbb{R}}f^+ dm \leq \liminf\int_{\mathbb{R}}f_{n_k}^+ dm$ and $\int_{\mathbb{R}}f^- dm \leq \liminf\int_{\mathbb{R}}f_{n_k}^- dm$. As $f_{n_k} \in S$, $\int_{\mathbb{R}}f_{n_k}^+\ dm = \int_{\mathbb{R}}f_{n_k}^-\ dm$, but that doesn't allow me to conclude that $\int_{\mathbb{R}}f^+\ dm = \int_{\mathbb{R}}f^-\ dm$.
I'm sure I'm overlooking something quite simple. Any hints on how to proceed would be very much appreciated.
This is the kernel of a continuous linear functional, hence closed. We have $|I(f_n)-I(f)|\leq\|f_n-f\|$, if $I(f_n)$ are all $0$ so is $I(f)$.