Let $\Omega\subset \mathbb{R}^N$ be a bounded, smooth domain. Assume that $u\in W_0^{1,1}(\Omega)$ and consider the distributional laplacean of $u$; $\Delta u$.
My question is: when is $\Delta u\in \mathcal{M} (\Omega)$?
If $\mu\in \mathcal{M}(\Omega)$ then, there is a unique $u\in W_0^{1,1}(\Omega)$ such that $-\Delta u=\mu$ in the distributional sense, so my question is equivalently to know if $$\mathcal{M}(\Omega)\subset\Delta(W_0^{1,1}(\Omega)),$$
is a strictly inclusion.
If $\Delta u$ is a signed measure, it can be decomposed into a positive and negative parts. Consequently, $u$ can be be written as a difference of two subharmonic functions. Such $u$ is called $\delta$-subharmonic.
There are $W^{1,1}$ functions that are not $\delta$-subharmonic. I'll give an example in one dimension: $u(x)=\int_0^x W(t)\,dt$ where $W$ is the Weierstrass function (continuous, nowhere differentiable). You can arrange for zero boundary values by subtracting a linear function. Clearly, $u\in C^1\subset W^{1,1}$. On the other hand, the second distributional derivative of $u$ is the first distributional derivative of $W$, which is an unimaginable horror. The set of unimaginable horrors is disjoint from the set of signed measures.
In one dimension, $u''$ is a signed measure iff $u'\in BV$. In higher dimensions, the condition $\nabla u\in BV$ is sufficient, but not necessary for $\Delta u$ to be a measure. The necessary and sufficient condition is for $\nabla u$ to be a divergence-measure field. (Look it up).