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The statement as printed is false without an integrability condition. On any nonempty open set, the constant functions $f_n\equiv n$ are harmonic and therefore subharmonic, but their supremum is $+\infty$, which is not locally integrable.

A valid version assumes $g={\sup }_n{f}_n\in L_{\mathrm{loc}}^1(\Omega )$. First, the maximum of finitely many subharmonic functions is subharmonic. For a distributional proof, mollify two functions on smaller interior domains and replace their maximum byThis is smooth, convex, and nondecreasing in each argument. The chain rule shows that its Laplacian applied to two smooth subharmonic functions is nonnegative: the first-derivative terms multiply their nonnegative Laplacians, and the Hessian term is nonnegative by convexity. Mollifier convergence in $L_{\mathrm{loc}}^1$ and the uniform bound $|M_{\delta }-\max |\le \delta /2$ allow both regularizations to be removed. Continuity of distributional derivatives under this convergence proves the finite-maximum assertion.

Now $g_N={\max }_{n\le N}{f}_n$ increases to $g$ and satisfies $|g_N|\le |f_1|+|g|$ almost everywhere. The dominated convergence theorem gives convergence in $L_{\mathrm{loc}}^1$. For every nonnegative test function,Thus $g$ is subharmonic in the distributional sense. This is the locally integrable supremum theorem for subharmonic functions. For the pointwise upper semicontinuous convention, take its canonical representative; the raw supremum need not already have that regularity.