Monotone Convergence Theorem:
Let ($f_n$) be a sequence in $\Sigma^+$ (i.e. measurable and nonnegative), such that $f_{n+1}\geq f_n$ almost everywhere for each $n$. Let $f=\limsup_n f_n$. Then $\mu (f_n)\uparrow \mu (f)\leq \infty$.
Can it be deduced that $f$ is measurable?
$f_n$ has to be measurable in order for
$$ \mu(f_n) = \int f_n d\mu $$
to be defined at all (at least for the usual definition of the integral).
Measurability of $f$ follows, because if each $f_n$ is measurable, you can show that the same is true of
$$ g_n := \sup_{\ell \geq n} f_\ell $$
for all $n$ (consider $g_n^{-1}((c,\infty])) = \bigcup_{\ell \geq n} f_\ell^{-1}((c,\infty))$), so that
$$ \limsup_n f_n = \lim_n \sup_{\ell \geq n} f_\ell $$
is also measurable (because pointwise limits of measurable functions are measurable).