In the book "Elliptic Partial Differential Equations of Second Order", the theorem 2.12 says the function $f(x)=\sup_{v\in S_\varphi}v(x)$ is harmonic in $\Omega$.
In the proof, it says there exists a sequence $\{v_n\}\in S_{\varphi}$ such that $v_n(y)\to u(y)$. By replacing $v_n$ with $\max(v_n,\inf\varphi)$, we may assume $v_n$ is bounded. $B=B_R(y)\subset\subset \Omega$, define $V_n$ to be the harmonic lifting of $v_n$. Then $V_n\in S_\varphi$ and $V_n(y)\to u(y)$
the harmonic lifting of $u$ is that $$ U(x)=\left\{ \begin{array}{l} \bar u(x),\quad x\in B\\ u(x),\quad x\in \Omega-B \end{array} \right . $$ $\bar u(x)$ is the harmonic function in $B$ given by the Poisson integral of $u$ on $\partial B$
So why $V_n(y)\to u(y)$ still holds?
Let $\Omega$ be bounded and $\varphi$ be a bounded function on $\partial \Omega$. A $C^0(\bar\Omega)$ subharmonic function $u$ is called a subfunction relative to $\varphi$ if it satisfies $u\leq \varphi$ on $\partial\Omega$.
$S_\varphi$ denote the set of subfunctions relative to $\varphi$.