Dominated convergence

Let $N_0$ be a measurable null set outside which $f_n(x)\to f(x)$ and $\mathrm{\mid }{f}_n(x)\mathrm{\mid }\le g(x)$ for every $n$, and let $N_{\mathrm{\infty }}:=\{g=\mathrm{\infty }\}$. By [L6], the set $N_{\mathrm{\infty }}$ is null. Put $N:=N_0\cup N_{\mathrm{\infty }}$, $E:=X\setminus N$, $\tilde{g}:=g{\chi }_E,h_n:=\mathrm{\mid }{f}_n-f\mathrm{\mid }{\chi }_E\mathrm{.}$ Then $h_n\to 0$ pointwise, $0\le h_n\le 2\tilde{g}$, and $\tilde{g}$ is nonnegative, measurable, and finite everywhere. Also $\mathrm{\mid }f\mathrm{\mid }\le \mathrm{\mid }f\mathrm{\mid }{\chi }_N+\tilde{g}\mathrm{.}$ By [L5] and [L7], $\int \mathrm{\mid }f\mathrm{\mid }d\mu \le \int \mathrm{\mid }f\mathrm{\mid }{\chi }_{N}d\mu +\int \tilde{g}d\mu =0+\int \tilde{g}d\mu \le \int gd\mu <+\mathrm{\infty },$ so $f\in L^1(\mu )$ by [L4]. The same null-set and domination argument gives $\int \mathrm{\mid }{f}_n\mathrm{\mid }d\mu \le \int gd\mu <\mathrm{\infty }$ for each $n$, so every $f_n$ also belongs to $L^1(\mu )$.

The functions $h_n$ are nonnegative, converge pointwise to $0$, and are dominated by the finite everywhere majorant $2\tilde{g}$. Applying [L1] therefore gives ${\mathrm{lim\hspace{0.17em}sup}}_n\int h_{n}d\mu \le \int 0d\mu =0.$ Hence $\int h_{n}d\mu \to 0$.

Because $\mathrm{\mid }{f}_n-f\mathrm{\mid }{\chi }_N$ is supported on the null set $N$, [L5] gives $\int \mathrm{\mid }{f}_n-f\mathrm{\mid }d\mu =\int h_{n}d\mu +\int \mathrm{\mid }{f}_n-f\mathrm{\mid }{\chi }_{N}d\mu =\int h_{n}d\mu ⟶0.$ Therefore, by [L2] and [L3], $\mid \int f_{n}d\mu -\int fd\mu \mid =\mid \int (f_n-f)d\mu \mid \le \int \mathrm{\mid }{f}_n-f\mathrm{\mid }d\mu ,$ and the right-hand side tends to $0$.