Minkowski's Integral Inequality when $p=\infty$ My version of the Minkowski's Integral Inequality is that, suppose $(X_1,\mathcal{M}_1,\mu_1)$ and $(X_2,\mathcal{M}_2,\mu_2)$ are complete $\sigma$-finite measure spaces and $f$ is non-negative measurable function on $X_1\times X_2$ with respect to the product measure $\mu_1\times \mu_2$, then for any $p\in [1,\infty]$ there is
$$\left\lVert \int f(x_1,x_2)dx_2\right\rVert_{L^p(X_1)}\leq \int \lVert f(x_1,x_2)\rVert_{L^p(X_1)}dx_2. $$
When $p=\infty $, this yields to prove that $\int f(x_1,x_2)dx_2\leq \int \lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}dx_2$ almost everywhere $x_1$, so (I think) we need to show that for almost everywhere $x_1$,
$$ f(x_1,x_2)\leq \lVert f(x_1,x_2)\rVert_{L^{\infty}(X_1)},\ \ a.e. x_2.$$
What we have by hand is that for any fixed $x_2\in X_2$, $f(x_1,x_2)\leq \lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}$ almost everywhere $x_1$, hence for any $x_2\in X_2$ there exists a nullset $Z(x_2)\subset X_1$ such that $f(x_1,x_2)\leq \lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}$ for any $x_1\in X_1\setminus Z(x_2)$. It follows that for $Z= \bigsqcup_{x_2\in X_2}Z(x_2)\times \{x_2\}\subset X_1\times X_2$ we have $f(x_1,x_2)\leq \lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}$ for any $(x_1,x_2)\in X_1\times X_2\setminus Z$. If we can show that $Z$ is measurable in $\overline{X_1\times X_2}$ the completion of $X_1\times X_2$, then by the Tonelli's Theorem $\overline{\mu_1\times \mu_2}(Z)=\int \mu_1(Z(x_2))dx_2=0$, whence for almost everywhere $x_1$ we have the inequality for almost everywhere $x_2$ as desired.
So the question reduces to show that $\bigsqcup_{x_2\in X_2}Z(x_2)\times \{x_2\}$ is measurable in the completion of $X_1\times X_2$ if each $Z(x_2)$ is of measure $0$. Is this true? If so, how to prove that? If not, then how to prove the Minkowski's Integral Inequality when $p=\infty$?
Thanks in advance.
 A: I figured out the solution after several discussions. For my argument, it remains only to show that the function $(x_1,x_2)\mapsto \lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}$ is measurable on $\overline{X_1\times X_2}$ and it suffices to show that $x_2\mapsto \lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}$ is measurable on $X_2$, so that the set $\{(x_1,x_2)\in X_1\times X_2\mid f(x_1,x_2)>\lVert f(x_1,x_2)\rVert_{L^\infty(X_1)} \}$ would be measurable and by "what we have by hand" we know that each of its slice has measure $0$, whence by Tonelli's Theorem this set is of measure $0$ and the conclusion follows. The condition that $x_2\mapsto \lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}$ is measurable on $X_2$ is in fact essential for this question, otherwise the integral $\int \lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}dx_2$ is never defined.
To show that $x_2\mapsto \lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}$ is measurable, observe that for any simple function $\varphi$ on $X_1\times X_2$, $\varphi\in L^P(X_1)$ for all $x_2\in X_2$ and any $1\leq p<+\infty$, it follows (trivially by direct calculation) that $\lVert \varphi(x_1,x_2)\rVert_{L^\infty(X_1)}=\lim_{p\to+\infty}\lVert \varphi(x_1,x_2)\rVert_{L^p(X_1)}$ for all $x_2$. Since by Tonelli's Theorem each $\lVert \varphi(x_1,x_2)\rVert_{L^p(X_1)}$ is measurable as a function on $X_2$, we know that $\lVert \varphi(x_1,x_2)\rVert_{L^\infty(X_1)}$ is measurable on $X_2$ for any simple $\varphi$. Now for any measurable $f$ on $X_1\times X_2$ with each of its slice $f(-,x_2)$ having finite measure support in $X_1$, choose a sequence of simple functions $\{\varphi_n\}$ that converges to $f$ pointwisely and Egorov's Theorem provides a simple proof for the fact that $\lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}=\lim_{n\to\infty}\lVert \varphi_n(x_1,x_2)\rVert_{L^\infty(X_1)}$ for all $x_2\in X_2$ as one can argue by contradiction. Thus $\lVert f(x_1,x_2)\rVert_L^\infty(X_1)$ is measurable in this case. To pass to general measurable functions $f$ on $X_1\times X_2$, write by $\sigma$-finite condition $X_1=\bigcup_{n=1}^\infty B_n$ with $B_n\subset B_{n+1}$ and $\mu_1(B_n)<\infty$ for all $n$. Observe again by contradiction that $\lVert f(x_1,x_2)\rVert_{L^\infty(X_1)}=\lim_{n\to\infty}\lVert f\cdot\chi_{B_n\times X_2}(x_1,x_2)\rVert_{L^\infty(X_1)}$ and the fact that each $\lVert f\cdot\chi_{B_n\times X_2}(x_1,x_2)\rVert_{L^\infty(X_1)}$ is measurable finishes the proof.
A: As far as I am concerned, it suffices to show that $$\sup_{x_1\in X_1}\int f(x_1,x_2)d\mu_2\le \int \sup_{x_1\in X_1}f(x_1,x_2)d\mu_2$$I think you can just use a kind of Fatou's Lemma(use the $\sup$, not $\inf$) to prove that.
