Proof. ($L^{p}$) $\Rightarrow$ (M): By Markov’s inequality.

($L^{p}$) $\Rightarrow$ (UI): Let $F \in \fF$ such that for each $f, g \in F$, $\norm{f - g}_{L^p(X; E)}< \eps$. Fix $f \in F$, then by the Dominated Convergence Theorem, there exists $M > 0$ such that $\int_{\bracs{\norm{f}_E \ge M}}\norm{f}_{E}^{p} d\mu < \eps$. For any $g \in F$,

so by Lemma 15.2.2,

Since $\bracs{\norm{g}_E \ge 2M, \norm{f}_E \le M}\subset \bracs{\norm{f - g}_E \ge M}$, by Markov’s inequality,

Therefore

($L^{p}$) $\Rightarrow$ (T): Let $F \in \fF$ such that for each $f, g \in F$, $\norm{f - g}_{L^p(X; E)}< \eps$. Fix $f \in F$, then by the Dominated Convergence Theorem, there exists $A \in \cm$ such that $\mu(A) < \infty$ and $\norm{\one_{A^c}f}_{L^p(X; E)}< \eps$. In which case, for any $g \in F$,

(M) + (UI) + (T) $\Rightarrow$ ($L^{p}$): Let $\eps > 0$. By (T), there exists $A \in \cm$ and $F_{1} \in \fF$ with $\mu(A) < \infty$ and $\sup_{f \in F_1}\int_{A^c}\norm{f}_{E}^{p} < \eps^{p}$. Thus for every $f, g \in F_{1}$,

By (UI), there exists $M > 0$ and $F_{2} \in \fF$ with $F_{2} \subset F_{1}$ such that

Assume without loss of generality that $\mu(A) > 0$ and let $\delta = \eps\mu(A)^{-1/p}$. By (M), there exists $F_{3} \in \fF$ with $F_{3} \subset F_{2}$, such that for any $f, g \in F_{3}$,

In which case,

then for any $f, g \in F_{3}$,

Now,

Similarly, $\normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}g}_{L^p(X; E)}\le 3\eps/2$. Thus

Therefore for any $f, g \in F_{3}$, $\norm{f - g}_{L^p(X; E)}\le 6 \eps$, so $\fF$ is Cauchy in $L^{p}(X; E)$.$\square$