 Solution

A simple function is a measurable function taking finitely many values, expressible as $s={\sum }_{j=1}^m{a}_j{1}_{A_j}$ for disjoint measurable sets. For nonnegative coefficients define $\int sd\mu ={\sum }_j{a}_j\mu (A_j)$. The Lebesgue integral of a nonnegative measurable $f$ isSuppose $0\le g_n\uparrow f$ with each $g_n$ simple. Their integrals increase and are bounded above by $\int f$. For the reverse bound, fix any simple $0\le s\le f$ and $0<c<1$. The sets $E_n=\{g_n\ge cs\}$ increase to a set containing $\{s>0\}$. Hence $\int g_n\ge c\int s{1}_{E_n}$, whose right side tends to $c\int s$ by continuity from below of the measure on the finitely many level sets of $s$. Let $c\uparrow 1$ and then take the supremum over $s$. This proves $\int g_n\uparrow \int f$, including infinite limiting integrals.

The monotone convergence theorem states the same conclusion for any nonnegative measurable sequence $f_n\uparrow f$, without simplicity. The preceding lower-bound argument works unchanged: $E_n=\{f_n\ge cs\}$ and monotonicity of the integral give $\int f_n\ge c\int s{1}_{E_n}$. Taking limits and the same two suprema provesThe upper bound follows from $f_n\le f$. Finiteness of the measure makes all bounded simple integrals finite in this proof; the theorem in fact extends to arbitrary measure spaces.