Let $\Omega$ be an open subset of $\mathbb{R}^n$ and let $u_k,\, u$ and $v$ be elements of $L^1_{\mathrm{loc}}(\Omega)$. Assume that \begin{equation} \begin{array}{ccc} u_k(x)\to u(x)\quad x\text{-a.e.} &\text{and}& u_k\to v\quad \text{in }\mathscr{D}'.\end{array} \end{equation} Does it follow that $u=v$ almost everywhere?
As it is usual, the convergence $u_k\to v$ in $\mathscr{D}'$ is defined as follows: $$\int_{\Omega}u_k(x) \phi(x)\, dx \to \int_{\Omega}v(x)\phi(x)\, dx,\qquad\text{for all } \phi \in C^{\infty}_c(\Omega).$$
The point of this question is to show that there is a minimum degree of "compatibility" between the two notions of convergence, even if neither of the two implies the other.
NOTE CAREFULLY. This question has already received an excellent answer by Chris Janjigian, which will be awarded the bounty.
Unless I have read this wrong, the answer is no. For example, consider $k\chi_{[0,\frac{1}{k}]}$. This sequence of functions converges pointwise to zero everywhere except at zero, but converges in the sense of distributions to the Dirac delta at zero.
Edit: Even if we assume that $v$ is in $L^1_{loc}$ the result is still not true. I am copying a large chunk of this answer from my solution to a problem in Rudin Real and Complex Analysis. The result proven here is that there exists a sequence of continuous functions converging to 0 pointwise $a.e.$, but converging to $1$ in the sense that $\langle g_n, \varphi \rangle \to \langle 1, \varphi \rangle$ for all $\varphi \in C[0,1]$.
To prove this, we will exploit the fact that Riemann integrals converge for continuous functions and, when they exist, coincide with Lebesgue integrals. It therefore suffices to show that we can produce a sequence measures which correspond to Riemann sums. The standard endpoint Riemann sum of mesh $\frac{1}{n^{2}}$ can be constructed by partitioning $[0,1]$ into intervals of length $\frac{1}{n^{n}}$ and, at the point $\frac{kn^{n-2}}{n^{n}}=\frac{k}{n^{2}}$, $1<k\leq n^{2}$ intervals placing a weighted Dirac delta $\frac{1}{n^{2}}\delta(x-\frac{k}{n^{2}})$ at the endpoint. Notice that the fact that we are using endpoint Dirac measure here does not actually matter since Riemann sums of continuous functions converge for any arbitrary point in the interval. We modify the above example slightly so that instead of using distributions we use continuous functions:
Keep the same partition, but in the interval $[\frac{kn^{n-2}-1}{n^{n}},\frac{k}{n^{2}}]$, draw a triangle (the explicit formula for this function is easy to produce and not terribly informative, so I exclude it) from $0$ to $2n^{n-2}$ and back down (with the vertex occurring at the midpoint). So $g_{n}$ is $0$ on $[\frac{k}{n^{2}},\frac{k+1}{n^{2}}-\frac{1}{n^{n}}]$ with a big spike just before every multiple of $\frac{1}{n^{2}}$. Integrating a function over this interval with respect to the measure induced by this function gives $\frac{1}{n^{2}}$ times a weighted average of the function over these intervals. This follows from continuity: for any fixed $f$ there exists $a,b\in[\frac{kn^{n-2}-1}{n^{n}},\frac{k}{n^{2}}]$ so that for all $x\in[\frac{kn^{n-2}-1}{n^{n}},\frac{k}{n^{2}}]$, $f(a)\leq f(x)\leq f(b).$ Then $$\frac{1}{n^{2}}f(a)\leq\int_{[\frac{kn^{n-2}-1}{n^{n}},\frac{k}{n^{2}}]}g_{n}fdm\leq\frac{1}{n^{2}}f(b)$$ and therefore there exists $x_{k}\in[\frac{kn^{n-2}-1}{n^{n}},\frac{k}{n^{2}}]$ so that this integral equals $\frac{1}{n^{2}}f(x_{k})$ by the mean value theorem. By construction, these functions integrate to $1$ (there are $n^{2}$ triangles each of which integrates to $\frac{1}{n^{2}})$. We can see from the above argument that for continuous functions the functional $f\to\int fg_{n}dm$ is in fact an operator sending $f$ to a Riemann sum of mesh $\frac{1}{n^{2}}$. Then for any continuous function the measures $g_{n}dm$ converge in the weak$^{*}$ sense to $m$. All that remains to be shown is that pointwise almost everywhere, these functions converge to $0$.
Define $A_{n}=\{x:g_{n}(x)>0\}$ and observe that $m(A_{n})=\frac{1}{n^{n-2}}$ (there are $n^{2}$ sets of measure $\frac{1}{n^{n}}$ on which $g_{n}$ is positive) which is summable. By the Borel Cantelli Lemma $\mu(\{x:g_{n}(x)\neq0\}\mbox{ for infinitely many $n$})=0$. It follows that $g_{n}\to0$ a.e. $\{m\}$.