TITLE: Compatibility of pointwise and distributional convergence
QUESTION [5 upvotes]: This has probably been asked before but I couldn't find it. 

Let $\Omega$ be an open subset of $\mathbb{R}^n$ and let $u_k,\, u \in L^1_{\mathrm{loc}}(\Omega)$ and $v\in \mathscr{D}'(\Omega)$ (the space of Schwartz's distributions). 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}
  Can we conclude that $v\in L^1_{\mathrm{loc}}(\Omega)$ and that $u=v$ almost everywhere?

Of course convergence in $\mathscr{D}'$ means that 
$$\int_{\Omega}u_k \phi\, dx \to \langle v, \phi\rangle\qquad\forall \phi \in C^{\infty}_c(\Omega).$$
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.

REPLY [6 votes]: 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 dense 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 occuring 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 fuction 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. $\{m\}$.