$H(\Omega)$ space of harmonic functions in $\Omega$. Show that $H(\Omega)$ is a closed subspace of $L^1(\Omega)$. > Let $\Omega$ be an open of $\mathbb{R}^N$ and $H(\Omega)$ the space constituted by the harmonic functions in $\Omega$ that belong to $L^1(\Omega)$. Show that $H(\Omega)$ is a closed subspace of $L^1(\Omega)$. If $f\in H_1(\Omega)$ then $\int_{\Omega} |f|<\infty$ and also $\Delta f = 0$ in $\Omega$. For $g\in H_1(\Omega)$, we have $f+g\in H_1(\Omega)$ because $\Delta(f+g) = \Delta f + \Delta g = 0$ and $\int_{\Omega} |f+g|\le \int_{\Omega} |f| + \int_{\Omega}|g| < \infty$ $\lambda f\in H_1(\Omega)$ because $\Delta \lambda f = \lambda \Delta f = 0$ and $\int_{\Omega}\lambda f = \lambda \int_{\Omega} f <\infty$ Is this it? I'm not so sure about the triangle inequality for integrals over regions like this. I know it works for integrals on one variable but why exactly can I use that? Why $\Omega$ must be open?

$\Omega$ must be open to define "harmonic". As Bongers said, you can do triangle inequality.

Suppose $f_n \to f$ in $L^1$, $f_n \in H(\Omega)$ for each $n$, and $f \in L^1(\Omega)$. By passing to a subsequence, we may assume $f_n \to f$ a.e.. Take $x \in \Omega$ and $r > 0$ such that $B_r(x) \subseteq \Omega$. Then $\int_{B_r(x)} f(y)dy = \lim_n \int_{B_r(x)} f_n(y)dy = \lim_n f_n(x) = f(x)$. Therefore $f$ is harmonic in $\Omega$, as desired.

There is a slight technicality with $f_n$ converging pointwise to $f$ almost everywhere, rather than everywhere. But I'll let you deal with this.