prove Minkowski's Inequality for Integrals 
Suppose that ($X$, $\mathcal{M}$, $\mu$) and ($Y$, $\mathcal{N}$, $\nu$) are $\sigma$-finite measure spaces, and let $f$ be an ($\mathcal{M} \otimes \mathcal{N}$)-measurable function on $X \times Y.$
a) If $f \ge 0$ and $1 \le p < \infty$, then
$$ \left[\int \left(\int f(x,y) d\nu(y) \right)^pd\mu(x)\right]^\frac{1}{p} \le \int \left[\int f(x,y)^p d\mu(x)\right]^\frac{1}{p}$$
b) If $1 \le p \le \infty$, $f(\cdot, y) \in L^p(\mu)$ for a.e. $y$, and the function $y \to ||f(\cdot, y)||_p$ is in $L^1(\nu)$, then $f(x, \cdot) \in L^1(\nu)$ for a.e. $x$, the function $x \to \int f(x,y) d\nu(y)$ is in $L^p(\mu)$, and $$\left|\left|\int f(\cdot, y)d\nu(y)\right|\right|_p \le \int||f(\cdot, y)||_pd\nu(y).$$  

 A: This is several year late, but here is another proof also based on Holder's inequality:
Without loss of generality we can assume that $f\geq0$.  The case $p=1$ is a restatement of Fubini's theorem. Suppose that $p>1$ and let  $H(x)=\int_Y f(x,y)\,\nu(dy)$. From Fubini's theorem and then H"older's inequality we obtain
\begin{aligned}
\|H\|^p_{L_p(\mu)} 
&=\int_X \int_Y f(x,y)\,\nu(dy) H^{p-1}(x)\,\mu(dx)
=\int_Y \int_X f(x,y) H^{p-1}(x)\,\mu(dx)\,\nu(dy)\\
&\quad \leq
\int_Y
\Big(\int_X|f(x,y)|^p\,\mu(dx)\Big)^{\tfrac{1}{p}}\|H\|^{p-1}_{L_p(\mu)}
\,\nu(dy).
\end{aligned}
The conclusion follows immediately if $\|H\|_p<\infty$.  If $\|H\|_p=\infty$, choose monotone sequences of sets $A_n\subset X$ and $B_n\subset Y$ such that $\mu(A_n)\vee\nu(B_n)<\infty$, and for any $k\in\mathbb{N}$ define $f_k=f\vee k$. Then
$$
\left(\int_{A_n}\Big(\int_{B_m}f_k(x,y)\,\nu(dy)\Big)^p\,\mu(dx)
\right)^{1/p}\leq
\int_{B_m}\Big(\int_{A_n}|f_k(x,y)|^p\,\mu(dx)\Big)^{1/p}\,\nu(dy).
$$
Letting first $k\rightarrow\infty$, then $n\rightarrow\infty$ and finally $m\rightarrow\infty$ we obtain the desired result.
A: Following Folland's proof (the inequality after applying Tonelli and Holder),  consider $\int f(x,y) \,dν(y)$ as a linear functional(not necessarily bounded) on $L_q(\mu)$. 
If it's bounded, then $\int f(x,y) \,dν(y)$ must be in $L_p(\mu)$ and the result is immediate.
Otherwise the RHS must be infinity. 
A: I am writing this answer just because it took me a while to understand the CQNKZX's answer.
I think that expanding it can help other people who, like me,were lost and really wanted to understand what is going on. So I decided to write a more detailed answer based entirely on the arguments of CQNZKX and Folland.
Let $p\in (1,\infty)$ and $q$ such that $1/p + 1/q = 1$. First of all, remember that the map
\begin{align*}
\Lambda: L^p(\mu) &\to L^q(\mu)^*\\
f &\mapsto \left(\Lambda(f)(g):= \int f(x)g(x)\  \mathrm{d} \mu\right)
\end{align*}
is an is an isometry with its image. Let $g\in L^q(\mu)$.
\begin{align}
\left|\Lambda\left(\int f(\cdot,y)\mathrm{d}\nu(y) \right)(g)\right|&= \int \left(\int f(x,y)\mathrm{d}\nu(y)\right)g(x) \mathrm{d}\mu(x)\\ 
&=\int \int f(x,y)g(x) \mathrm{d}\mu(x)\mathrm{d}\nu(y) \ \ \ \ \ \ \ \ \ \quad (1) \\
&\leq \int \left \Vert  f(\cdot,y) \right\Vert_p \|g\|_q\ \mathbb{d} \nu(y) \ \ \ \  \ \ \ \ 
 \quad \quad \ \  \quad (2)\\
&\leq \|g\|_q \int\|f(\cdot,y)\|_p \ \mathrm{d} \nu(y).
\end{align}
In (1) we have used Fubinni's theorem and in (2) Holder's inequality.
Since $\Lambda$ is an isometry and the above equality holds for every $g\in L^q(\mu)$
$$\left\|\int f(\cdot,y)\  \mathrm{d}\nu(y) \right\|_p = \left\| \Lambda\left( \int f(\cdot,y)\  \mathrm{d}\nu(y) \right)\right\|_{\mathrm{operator}} \leq \int\|f(\cdot,y)\|_p \ \mathrm{d} \nu(y).$$
If $p=1$ the inequality is trivial.
