Let $u$ be a subharmonic function on an open set $U$ in $\mathbb{C}$, and let $v$ be an upper semicontinuous function on $U$ such that $u\leq v$ almost everywhere. Prove that $u\leq v$ everywhere.
Suppose $u\leq v$ on $U-E$ where $E$ has measure zero. If $v$ were also subharmonic then we can show the result by using Smoothing Theorem which focuses on convolutions. So I tried to prove that $v$ has submean property on $U$, but failed. Then I tried to get a contradiction by assuming that there is $\omega_0\in E$ such that $u(\omega_0)>v(\omega_0)$, but have no idea how to proceed from here. Any suggestion please..
Let $a \in U$ and $\varepsilon > 0$. From the upper semi-continuity of $v$ it follows that there is an $R > 0$ such that $$ v(z) < v(a) + \varepsilon \text{ for } \lvert z - a \rvert \le R \, . $$
$u$ is subharmonic. Integrating the sub-mean inequality at $z=a$ for $0 \le r \le R$ gives $$ u(a) \le \frac{1}{ \pi R^2 } \int_0^{2 \pi} \int_0^R u(a + r e^{it}) \, r \, dr \, dt $$ Since $u(z) \le v(z)$ almost everywhere in $U$ it follows that
$$ u(a) \le \frac{1}{ \pi R^2 } \int_0^{2 \pi} \int_0^R v(a + r e^{it}) \, r \, dr \, dt \le v(a) + \varepsilon \, . $$ This holds for arbitrary $\varepsilon > 0$, therefore $u(a) \le v(a)$.