Proof of Dominated Convergence Theorem

Claim 1. Since $f(x)=\lim_m f_m(x)$ for every $x$, we have $f=\liminf_m f_m$ pointwise, where for real sequences $\liminf_m a_m=\sup_k\inf_{m\ge k}a_m$; here all values lie in $[-g(x),g(x)]$, so the infima and suprema are real. Measurability of $f$ follows exactly as in the proof of Fatou's Lemma: the functions $x\mapsto\inf_{m\ge k}f_m(x)$ are measurable via $\{\cdot\ge a\}=\bigcap_{m\ge k}\{f_m\ge a\}$ and the criterion of that definition, and their pointwise supremum over $k$ is measurable via $\{\cdot>a\}=\bigcup_k\{\cdot_k>a\}$. Moreover $|f|\le g$ pointwise (limits preserve weak inequalities), so $\int_X|f|\,d\mu\le\int_X g\,d\mu<\infty$ by monotonicity (claim 1 of Linearity and Monotonicity of the Lebesgue Integral) and $f$ is integrable.

Claim 2. Set $h_m=2g-|f_m-f|$. Each $h_m$ is measurable (differences and absolute values of measurable functions are measurable, by Step 0(a) of the proof of Linearity and Monotonicity of the Lebesgue Integral together with $|u|=u^{+}+u^{-}$ and the measurability of positive and negative parts from Integrable Function and the Lebesgue Integral), and $h_m\ge 0$ pointwise since $|f_m-f|\le|f_m|+|f|\le 2g$. Also $h_m(x)\to 2g(x)$ for every $x$, so $\liminf_m h_m=2g$ pointwise. By Fatou's Lemma,

where $\limsup_m a_m=\inf_k\sup_{m\ge k}a_m$ for a bounded real sequence, and the last equality uses the linearity of the integral for the integrable functions $2g$ and $|f_m-f|$ (claim 2 of Linearity and Monotonicity of the Lebesgue Integral) together with the elementary identity $\liminf_m(C-a_m)=C-\limsup_m a_m$ for real sequences and constants. Since $\int_X 2g\,d\mu<\infty$, subtracting it gives $\limsup_m\int_X|f_m-f|\,d\mu\le 0$; the terms are nonnegative, so $\int_X|f_m-f|\,d\mu\to 0$.

Claim 3. By claim 2 of Linearity and Monotonicity of the Lebesgue Integral,

so $\int_X f_m\,d\mu\to\int_X f\,d\mu$ in the sense of Limit of a Sequence of Real Numbers. $\blacksquare$