Weak lower semicontinuity of an integral of a convex function Let $f:\mathbb{R} \to \mathbb{R}$ be a convex function. Is there an easy way to show that the map
$$u \mapsto \int_{\Omega}f(u(x))\;dx$$
is weakly lower semicontinuous for $u \in L^2(\Omega)$? (Suppose that $f(u)$ is in whatever $L^p$ space is required).
I would prefer a proof without going into the epigraph terminology, which is the only way that I have seen it.
 A: On sets of infinite measure, even the strong  $L^2$ convergence does not imply much for $\int f(u)$. For example, take $u_n = -\frac1n \chi_{[0,n]}$; then  $u_n\to 0$ in $L^2(\mathbb R)$ but $\int u_n =-1< 0$. So, I am going to assume that $\Omega$ has finite measure. 
Fix $u$ and $\epsilon>0$. There is $M$ such that 
$$  \int_{|u|\ge M}  f\circ u   < \epsilon$$
Partition $[-M,M) $ into finitely many disjoint halfopen intervals $I_k$ such that for each $k$ there is an affine function $g_k(x)=a_kx+b_k$ such that $g_k\le f$ on $\mathbb R$ and $g_k\ge f-\epsilon $ on $I_k$. 
Suppose $u_n\to u$ weakly in $L^2$ (or in $L^1$, which is enough). Then $$\int_{\Omega} (au_n+b) \to \int_{\Omega} (au+b) $$ for every $a,b\in \mathbb R$. By subtracting an affine function (e.g., support function at $0$) from $f$, we may assume $f\ge 0$ on $\mathbb R$. 
For each $k$ we have 
$$\int_{\Omega_k} f\circ u_n \ge \int_{\Omega_k} (a_ku_n+b_k) \to  \int_{\Omega_k} (au+b) \ge 
\int_{\Omega_k} (f\circ u -\epsilon)
$$
Sum over $k$ to obtain 
$$\liminf\int_{\Omega} f\circ u_n \ge
\liminf\int_{|u|\le M} f\circ u_n \ge \int_{|u|\le M}( f\circ u -\epsilon)$$
Since $\epsilon$ was arbitrarily small, lower semicontinuity follows.
