Vitali convergence theorem on finite and sigma-finite measure spaces

Assume $\mu (X)<+\mathrm{\infty }$ and $f_n\to f$ in $L^1(\mu )$. By [L4], the sequence converges to $f$ in measure. To prove uniform integrability, it suffices by [L3] to show $L^1$-boundedness and uniform absolute continuity. For $L^1$-boundedness, choose $N$ so that $\int \mathrm{\mid }{f}_n-f\mathrm{\mid }d\mu <1$ for $n\ge N$. Then $\int \mathrm{\mid }{f}_n\mathrm{\mid }d\mu \le \int \mathrm{\mid }f\mathrm{\mid }d\mu +1$ for $n\ge N$, and the finitely many initial terms may be absorbed into one larger bound.

Still on a finite measure space, assume $f_n\to f$ in measure and the family $\{f_n\}$ is uniformly integrable. By [L3], the family is $L^1$-bounded and uniformly absolutely continuous on small sets. By [L5], some subsequence $f_{n_k}$ converges to $f$ almost everywhere. Fatou's lemma [L7] then gives $\int \mathrm{\mid }f\mathrm{\mid }d\mu \le {\mathrm{lim\hspace{0.17em}inf}}_{k\to \mathrm{\infty }}\int \mathrm{\mid }{f}_{n_k}\mathrm{\mid }d\mu <+\mathrm{\infty },$ so $f\in L^1(\mu )$.

Still under step 1.1, let $\epsilon >0$. Choose $N$ so that $\int \mathrm{\mid }{f}_n-f\mathrm{\mid }d\mu <\epsilon \mathrm{/}2$ for $n\ge N$. Apply [L6] to $f$ and to the finitely many functions $f_0,\dots ,f_{N-1}$, and let $\delta$ be the minimum of the resulting positive numbers. Then $\mu (E)<\delta$ implies ${\int }_E\mathrm{\mid }{f}_n\mathrm{\mid }d\mu <\epsilon$ for every $n<N$, while for $n\ge N$ one has ${\int }_E\mathrm{\mid }{f}_n\mathrm{\mid }d\mu \le {\int }_E\mathrm{\mid }f\mathrm{\mid }d\mu +\int \mathrm{\mid }{f}_n-f\mathrm{\mid }d\mu <\epsilon \mathrm{.}$ So [L3] yields uniform integrability.

Let $\epsilon >0$. Choose $\delta >0$ such that $\mu (E)<\delta$ implies both ${\sup }_n{\int }_E\mathrm{\mid }{f}_n\mathrm{\mid }d\mu <\epsilon \mathrm{/}3$ and ${\int }_E\mathrm{\mid }f\mathrm{\mid }d\mu <\epsilon \mathrm{/}3$; the first uses step 1.2 together with [L3], and the second uses [L6]. Choose $\eta >0$ with $\eta \mu (X)<\epsilon \mathrm{/}3$. Since $f_n\to f$ in measure, there is $N$ such that $\mu (B_n)<\delta$ for $n\ge N$, where $B_n:=\{\mathrm{\mid }{f}_n-f\mathrm{\mid }>\eta \}$. Then for $n\ge N$, $\int \mathrm{\mid }{f}_n-f\mathrm{\mid }d\mu \le {\int }_{B_n}\mathrm{\mid }{f}_n\mathrm{\mid }d\mu +{\int }_{B_n}\mathrm{\mid }f\mathrm{\mid }d\mu +{\int }_{X\setminus B_n}\eta d\mu <\epsilon \mathrm{.}$ So $f_n\to f$ in $L^1(\mu )$.

Assume now that $\mu$ is sigma-finite, that $f_n\to f$ in measure, and that the family $\{f_n\}$ is uniformly integrable and tight. Let $\epsilon >0$. By [L8], choose a measurable set $E$ with $\mu (E)<+\mathrm{\infty }$ and ${\sup }_n{\int }_{X\setminus E}\mathrm{\mid }{f}_n\mathrm{\mid }d\mu <\epsilon \mathrm{/}4.$ Because $f_n\to f$ in measure on $X$, the restricted sequence converges in measure on $E$ as well, and uniform integrability persists on $E$. Therefore step 2.2 applied on the finite measure space $E$ gives $f{\mathrm{\mid }}_E\in L^1(\mu )$ and ${\int }_E\mathrm{\mid }{f}_n-f\mathrm{\mid }d\mu \to 0$. By [L5], after passing to a subsequence if necessary we may assume $f_{n_k}\to f$ almost everywhere on $X$. Applying [L7] to the nonnegative functions $\mathrm{\mid }{f}_{n_k}\mathrm{\mid }{\mathbf{1}}_{X\setminus E}$ gives ${\int }_{X\setminus E}\mathrm{\mid }f\mathrm{\mid }d\mu \le {\mathrm{lim\hspace{0.17em}inf}}_{k\to \mathrm{\infty }}{\int }_{X\setminus E}\mathrm{\mid }{f}_{n_k}\mathrm{\mid }d\mu \le \epsilon \mathrm{/}4.$ Hence $f\in L^1(\mu )$. For all large $n$ one has ${\int }_E\mathrm{\mid }{f}_n-f\mathrm{\mid }d\mu <\epsilon \mathrm{/}2$, and then $\int \mathrm{\mid }{f}_n-f\mathrm{\mid }d\mu \le {\int }_E\mathrm{\mid }{f}_n-f\mathrm{\mid }d\mu +{\int }_{X\setminus E}\mathrm{\mid }{f}_n\mathrm{\mid }d\mu +{\int }_{X\setminus E}\mathrm{\mid }f\mathrm{\mid }d\mu <\epsilon \mathrm{.}$ So $f_n\to f$ in $L^1(\mu )$.

Step 1.1 proves the forward finite implication, steps 1.2 and 2.2 prove the reverse finite implication, and step 3.1 proves the sigma-finite tight form.