Can we conclude that $\Delta (\Phi\circ u)$ is a measure, given that $\Phi$ is a particular smooth function and $u$ is in some Sobolev space? Let $\Omega\subset\mathbb{R}^N$ be a bounded smooth domain. Assume that $u\in W_0^{1,2}(\Omega)$ is such that $\Delta u$ is a measure. 
Let $\Phi$ be a smooth function in $\mathbb{R}$, such that $\Phi'(u)\in W_0^{1,2}(\Omega)$. Is it true that $\Delta (\Phi\circ u)$ is a measure and $$\Delta (\Phi\circ u)=(\Phi''\circ u)|\nabla u|^2+(\Phi'\circ u)\Delta u, \tag{1}$$
If the equality $$\int_\Omega \phi \Delta u=-\int_\Omega \nabla \phi \nabla u,\ \forall \phi\in W_0^{1,2}(\Omega),\tag{2}$$
were true, then I can show that $(1)$ is true, however, I fail to see if $(2)$ is true. Any idea is appreciated.
Remark: I don't know if it matter, but in my particular case, I have that $u$ is superharmonic also.
Remark 2: Equality $(2)$ is true for all $\phi\in C_0^\infty(\Omega)$.
Remark 3: If $(2)$ is true, I think the proof goes as follows: if $\varphi\in C_0^\infty(\Omega)$ then (remember that $\Phi'\circ u\in W_0^{1,2}(\Omega)$, so that $\varphi(\Phi'\circ u)\in W_0^{1,2}(\Omega)$)
\begin{eqnarray}
 \int_\Omega (\Phi''\circ u)|\nabla u|^2\varphi+\int_\Omega \varphi(\Phi'\circ u)\Delta u &=& \int_\Omega (\Phi''\circ u)|\nabla u|^2\varphi-\int_\Omega \nabla (\varphi(\Phi'\circ u))\nabla u    \nonumber \\
   &=& \int_\Omega \varphi\nabla (\Phi'\circ u)\nabla u-\int_\Omega\varphi\nabla (\Phi'\circ u)\nabla u-\\ 
&& \int _\Omega (\Phi'\circ u)\nabla u\nabla\varphi \nonumber \\
   &=& \int_\Omega \nabla (\Phi\circ u)\nabla \varphi = \int_\Omega (\Phi\circ u)\Delta \varphi
\end{eqnarray}
therefore $(1)$ is true.
Update: Motivated by @thisismuch answer, we will include the additional hypothesis: $u\ge 0$, $u$ is superharmonic and there are $0<k<t<1$ such that $\Phi(x)=0$ for $x\le k$ and $\Phi(x)=1$ for $x\ge t$.
Note that $|\Delta u|=-\Delta u$, therefore
\begin{eqnarray}
 \left|\int \nabla u\nabla ((\Phi'\circ u)\phi)\right| &=& \left|\int_{spt(\phi)\cap \{u>k\}}\nabla u\nabla ((\Phi'\circ u)\phi) \right|     \nonumber \\
   &=& \left|\int_{spt(\phi)\cap \{u>k\}}\phi (\Phi'\circ u)\Delta u\right| \nonumber \\
   &\le& c\|\phi\|_\infty \int_{spt(\phi)\cap \{u>k\}}(-\Delta u) \\
&\le & c\|\phi\|_\infty\int_{spt(\phi)} \frac{u(-\Delta u)}{k}
\end{eqnarray}
I would like to apply Green identity in the last equality. Is the following equality true? $$\int_\Omega u\Delta u=-\int_\Omega |\nabla u|^2$$
 A: I can show (1), but I don't know if $\Delta(\Phi\circ u)$ is a measure.
The equality (1) has to be understood in the sense of distributions, so we must figure out how $(\Phi'\circ u)\Delta u$ acts on test functions. We know how $\Delta u$ acts: $T(\phi)=\int u\Delta \phi$. Since $u$ is a $W^{1,2}$ function, $T$ is a particularly nice distribution: its value on $\phi$ is controlled by the $W^{1,2}$ norm of $\phi$: 
$$\left|T(\phi)\right| \le \int |\nabla u \nabla \phi|\le \|\nabla u\|_{L^2} \|\nabla \phi\|_{L^2}\tag{A}$$  This allows us to extend $T$ to a continuous linear functional on $W^{1,2}_0$. In fact, this extension is simply 
$$T(\phi) =  -\int  \nabla u \nabla \phi,\quad \phi\in W^{1,2}_0 \tag{B}$$
because both sides are continuous on $W^{1,2}_0$ and agree on a dense subset. 
Now we can make sense of $(\Phi'\circ u)\Delta u$: it is the distribution $$\phi \mapsto T((\Phi'\circ u) \phi)\tag{C} $$
which is defined because $(\Phi'\circ u) \phi\in W_0^{1,2}$.
Assuming $\Phi''$ is bounded, the term $(\Phi''\circ u)|\nabla u|^2$ in (A) is an $L^1$ function.  Now that the right hand side of (1) makes sense, we can apply it to test function $\phi$ and  find, using (B)-(C) (as in your Remark 3) that it results in $\int (\Phi\circ u)\Delta \phi$. This qualifies the right hand side of (1) as the distributional Laplacian of $\Phi\circ u$. 
But is it a measure? For a distribution to be a measure, it must be controlled by the supremum norm of  test functions. The first term on the right in (1) is an $L^1$ function, no problem there. The problem is the second term. We need
$$
\left|\int \nabla u\nabla \left((\Phi'\circ u) \phi\right)\right\| \lesssim \sup |\phi|
\tag{D}$$
and I don't see why this would be true.
