Monotony is a superfluous hypothesis in the Monotone convergence theorem for Lebesgue integral. In fact the following is true.
Theorem - Let $(X, \tau, \mu)$ be a measurable space, $f_n : X \rightarrow [0,\infty]$ a sequence of measurable functions converging almost everywhere to a function $f$ so that $f_n \leq f$ for all $n$. Then $$\lim_{n\rightarrow \infty} \int_X f_n d\mu = \int_X f d\mu.$$
Proof: $$ \int_X f d\mu = \int_X \underline{\lim} \, f_n d\mu \leq \underline{\lim} \int_X f_n d\mu \leq \overline{\lim} \int_X f_n d\mu \leq \int_X f d\mu. $$
I learnt this result from an article by J.F. Feinstein in the American Mathematical Monthly, but I never saw it in any textbook. Since the Monotone convergence theorem is important, I wish to argue that this is also an important theorem. Here is an illustration.
Let $(X, \tau, \mu)$ be a measurable space, $f : X \rightarrow [0,\infty]$ a measurable function. Then $$ \int_X f d\mu = \lim_{r \rightarrow 1, r>1} \sum_{n\in {\bf Z}} r^n \mu\Bigl( f^{-1}([r^n, r^{n+1}))\Bigr). $$ Neither the dominated nor the monotone convergence theorem apply here. Note that this is a way to define the Lebesgue integral of nonnegative functions. Computing integrals by geometrically dividing the $x$ axis is due to Fermat.