Open subgroup and group algebra

Question

Let $G$ be a locally compact group and $H$ be an open subgroup of $G$. Consider the group algebras $L^1(G)$ and $L^1(H)$ with convolution product and consider $L^1(H)$ as a subalgebra of $L^1(G)$ (i.e., $L^1(H)=\{f\in L^1(G): f|_{G\setminus H}=0\}$). Prove that $L^1(H)$ is closed subalgebra of $L^1(G)$.

Answer 1

Choose $f\in\overline{L^1(H)}\subset L^1(G)$. There exists a sequence $\{f_n\}_n\subset L^1(H)$ such that $f_n\to f$ in $L^1(G)$. We know from real analysis this concludes there exists a subsequence $\{f_{n_k}\}_k$ of $\{f_n\}_n$ such that $f_{n_k}\to f$ almost every where. Hence except a set, of measure zero for all $x$ $f_{n_k}(x)\to f(x)$. But we know $f_{n_k}(x)=0$ when $x\in G\setminus H$ and so $f=0$ almost every where on $G\setminus H$ which means $f\in L^1(H)$.