Minkowski's integral inequality

If $p=1$, then Tonelli directly gives [L2, given, algebra] $\mathrm{\parallel }H{\mathrm{\parallel }}_{L^1(X)}={\int }_X{\int }_Y\mathrm{\mid }F(x,y)\mathrm{\mid }d\nu (y)d\mu (x)={\int }_Y\mathrm{\parallel }F(\cdot ,y){\mathrm{\parallel }}_{L^1(X)}d\nu (y)\mathrm{.}$

Assume $1<p<\mathrm{\infty }$ and let $q$ be conjugate to $p$. Put [L2, L3, given, algebra] $M:={\int }_Y\mathrm{\parallel }F(\cdot ,y){\mathrm{\parallel }}_{L^p(X)}d\nu (y)\mathrm{.}$ For every nonnegative $g\in L^q(X)$ with $\mathrm{\parallel }g{\mathrm{\parallel }}_q\le 1$, ${\int }_{X}H(x)g(x)d\mu (x)={\int }_Y({\int }_X\mathrm{\mid }F(x,y)\mathrm{\mid }g(x)d\mu (x))d\nu (y)$ by [L2], and then [L3] yields ${\int }_X\mathrm{\mid }F(x,y)\mathrm{\mid }g(x)d\mu (x)\le \mathrm{\parallel }F(\cdot ,y){\mathrm{\parallel }}_p\mathrm{\parallel }g{\mathrm{\parallel }}_q\le \mathrm{\parallel }F(\cdot ,y){\mathrm{\parallel }}_p\mathrm{.}$ Hence ${\int }_{X}Hgd\mu \le {\int }_Y\mathrm{\parallel }F(\cdot ,y){\mathrm{\parallel }}_{p}d\nu (y)=M\mathrm{.}$

Choose measurable sets $X_1\subseteq X_2\subseteq \cdots$ with [given, construct] $\mu (X_m)<\mathrm{\infty }$ and ${\bigcup }_m{X}_m=X$, and define $H_m:=\min (H,m){\mathbf{1}}_{X_m}\mathrm{.}$ Then $0\le H_m\uparrow H$ pointwise, and each $H_m$ lies in $L^p(X)$ because $H_m\le m{\mathbf{1}}_{X_m}$.

Applying [L1] to each nonnegative $H_m$ and using $0\le H_m\le H$ with [L1, step 1.2, step 1.3] step 1.2 gives $\mathrm{\parallel }{H}_m{\mathrm{\parallel }}_p\le M$ for every $m$.

By [L4], ${\int }_X{H}_m(x{)}^{p}d\mu (x)\uparrow {\int }_{X}H(x{)}^{p}d\mu (x)\mathrm{.}$ Since $\int H_m^{p}d\mu =\mathrm{\parallel }{H}_m{\mathrm{\parallel }}_p^p\le M^p$ for every $m$, the limit is finite and satisfies $\mathrm{\parallel }H{\mathrm{\parallel }}_p^p\le M^p$. Hence $H\in L^p(X)$ and $\mathrm{\parallel }H{\mathrm{\parallel }}_p\le M={\int }_Y\mathrm{\parallel }F(\cdot ,y){\mathrm{\parallel }}_{L^p(X)}d\nu (y)\mathrm{.}$ Together with step 1.1, this proves the theorem for all $1\le p<\mathrm{\infty }$.