I hope it's OK to post with a proof rather than a reference; when I started writing I thought this would be shorter...

**Lemma.** Let $(f_n \mid n \geq 1)$ be a sequence of convex functions $f_n: \mathbf{R}^d \to \mathbf{R}$, assumed to converge a.e. to a function $f: \mathbf{R}^d \to \mathbf{R}$. Then (i) the convergence is locally uniform, and (ii) $f$ is convex.

Throughout let $D_R \subset \mathbf{R}^d$ be the $d$-dimensional disc of radius $R$. We will use the three nested discs $D_1 \subset D_2 \subset D_5$, ultimately showing that the convergence is uniform on the innermost one: we prove first (in Claim 1) that the $f_n$ are bounded above on $D_5$, next (in Claim 2) that they are bounded below on $D_2$, and finally (in Claim 3) that they are equicontinuous on $D_1$. The conclusion follows by Arzela-Ascoli.

Recall first that convex functions are continuous (on the interior of their domains); this is true in particular for every $f_n$.

For all points $x \in \mathbf{R}^d$ and unit vectors $v \in \mathbf{R}^d$, let $\mathbf{L}_{x,v} \subset \mathbf{R}^d$ be the affine line through $x$ directed by $v$. For example, let $x \neq y \in \mathbf{R}^d$, and $v = y - x/\lvert y - x \rvert$. Any convex function $h: \mathbf{R}^d \to \mathbf{R}$ remains convex when restricted to $\mathbf{L}_{x,v}$, and along this line one has $h(x+tv) \geq h(x) + t(h(y)-h(x))/\lvert y - x \rvert$ for all $t \not\in (0,\lvert y - x \rvert)$. We will use this repeatedly with $h = f_n$.

*Claim 1.* There is $M > 0$ so that $f_n \leq M$ in $D_5$ for all $n$.

*Proof.* If not, then there is a sequence of points $x_n \in D_3$ along which $M_n := f_n(x_n) \to + \infty$. Let $\mathbf{L}_{x_n,v}$ be an arbitrary line through $x_n$. Along it, $f_n(x_n + tv) \geq f_n(x_n) + t (f_n(x_n + v) - f_n(x_n))$ for all $t \not \in (0,1)$. Pick $t < -1$ or $t > 1$ depending on the sign of $f_n(x_n + v) - f_n(x_n)$. For one of the half-lines one has $f_n \geq M_n$ along it. By varying $v$ one finds that $f_n \geq M_n$ on at least *half* of the annulus $D_2(x_n) \setminus D_1(x_n)$: $\mathcal{H}^d( \{ x \in D_2(x_n) \setminus D_1(x_n) \mid f_n(x) \geq M_n \}) \geq \mathcal{H}^d( D_2 \setminus D_1) / 2$. This is absurd because $D_2(x_n) \subset D_5$, and $f_n$ converges a.e. to $f$ in that set. $\blacksquare$

*Claim 2.* There is $C > 0$ so that $f_n \geq - CM$ in $D_2$ for all $n$.

*Proof.* We argue by contradiction: as $f_n \leq M$ on $D_5$, if there were a sequence of points $x_n \in D_2$ with $f_n(x_n) \leq -C_n M_n \to \infty$, then $f_n \to -\infty$ on $D_2$. But this is impossible because $f_n$ converges a.e. in $D_2$.

Take thus some fixed $C > 0$, some sequence $C_n \to +\infty$, and two sequences of points $x_n,y_n \in D_2$ so that $f_n(x_n) = -C_n M$ but $f_n(y_n) \geq -CM.$

Let $v_n = y_n - x_n / \lvert y_n - x_n \rvert$. Along the line $\mathbf{L}_{x_n,v_n}$,
\begin{align}
f(x_n + t v_n)
&\geq f(x_n) + t(f(y_n) - f(x_n))/\lvert y_n - x_n \rvert \\
&\geq -C_n M + t(-CM + C_n M)/\lvert y_n - x_n \rvert \\
&\geq MC_n(-1 + t(1 - C/C_n)/\lvert y_n - x_n \rvert)
\end{align}
for all $t > \lvert y_n - x_n \rvert$. We may assume that $C_n/100 > C \geq 1$ say; therefore if $t \in [2 + \lvert y_n - x_n \rvert, 3 + \lvert y_n - x_n \rvert]$ then
\begin{equation}
f(x_n + t v_n)
\geq MC_n(-1 + 2 (1 - C/C_n)) \geq 99M.
\end{equation}
But for this range of $t$, $x_n + t v_n \in D_5$, where $f_n \leq M$; this is absurd. $\blacksquare$

*Claim 3.* The $f_n$ are $2CM$-Lipschitz in $D_1$.

*Proof.*
We argue again by contradiction. Let $x_n,y_n \in D_1$ be two sequences of points along which $f_n(y_n) \geq f_n(x_n)$ and $f_n(y_n) - f_n(x_n) > 2CM \lvert y_n - x_n \rvert$. Let $v_n$ be the unit vector from $x_n$ to $y_n$. Along the line $\mathbf{L}_{x_n,v_n}$,
\begin{equation}
f_n(x_n + tv_n) > f_n(x_n) + 2tCM \geq CM
\end{equation}
for all $t \in [1 + \lvert y_n - x_n \rvert,2+\lvert y_n - x_n \rvert]$. This is absurd because $x_n + t v_n \in D_5$ for this range of $t$, where $f_n \leq M$. $\blacksquare$