The <Beppo Levi theorem>, or <monotone convergence theorem>, states that if $(f_n)$ is a sequence of nonnegative measurable functions on a <measure space> and
$$
f_n\uparrow f
$$
pointwise almost everywhere, then $f$ is measurable and
$$
\boxed{\int f\,d\mu
=\lim_{n\to\infty}\int f_n\,d\mu},
$$
where either side may be infinite.

Measurability of $f=\sup_nf_n$ follows from measurability of countable suprema. Since $f_n\leq f$, <monotonicity of the Lebesgue integral> gives
$$
\lim_n\int f_n\,d\mu\leq\int f\,d\mu.
$$
For the reverse inequality, let $s$ be a nonnegative <simple function> with $s\leq f$, and fix $0<c<1$. Define
$$
E_n=\{x:f_n(x)\geq c,s(x)\}.
$$
Then $E_n\uparrow X$ up to the null set on which convergence fails. Since $f_n\geq c,s\mathbf1_{E_n}$,
$$
\int f_n\,d\mu
\geq c\int_{E_n}s\,d\mu
\longrightarrow c\int s\,d\mu
$$
by <continuity from below of a measure>, applied to the finitely many level sets of $s$. Hence $\lim_n\int f_n\geq c\int s$. Letting $c\uparrow1$ and taking the supremum over all simple $s\leq f$, as in the definition of the <Lebesgue integral>, gives the reverse inequality and proves the theorem.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 23H / d / Solution

{wiki}

The Lebesgue integral of a nonnegative measurable function is the supremum of the integrals of nonnegative simple functions below it. Integrable real and complex functions are then defined from their positive, negative, real, and imaginary parts.


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