Suppose $(S_i,\F_i,\mu_i)$, $i=1,2$, are measure spaces. If $f:S_1\rar L_1(S_2,\mu_2)$ is integrable, then there is some function $F\in L_1(S_1\times S_2,\mu_1\otimes\mu_2)$, such that
  1. For $\mu_1$ almost all $x_1$: $F_{x_1}=f(x_1)$.
  2. For $\mu_2$ almost all $x_2$: $F_{x_2}$.
Put $E\colon=L_1(\O_1,\F_1,\mu_1,L_1(\mu_2))$. If $f$ is simple, then $$ F=\sum_{j=1}^n g_jI_{A_j} \quad\mbox{where}\quad g_j\in L_1(\mu_2), $$ Now define $$ F(x_1,x_2)\colon=\sum_{j=1}^n g_j(x_2)I_{A_j}(x_1)~. $$ We have: $F_{x_1}=f$ and $\int F_{x_2}\,d\mu_1=(\int f\,d\mu_1)(x_2)$.
If $f$ is integrable take a determining sequence $f^n$ and the corresponding sequence $F^n$. By Fubini we get \begin{eqnarray*} \norm{F^n-F^m}_{L_1(\mu)} &=&\iint|F^n_{x_1}-F^m_{x_1}|\,d\mu_2\,\mu_1(dx_1)\\ &=&\int\tnorm{f^n(x_1)-f^m(x_1)}_{L_1(\mu_2)}\,\mu_1(dx_1) =\norm{f^n-f^m}_E \end{eqnarray*} Thus $F^n$ is Cauchy in $L_1(\mu)$; hence it converges to $F\in L_1(\mu)$. We know by Fubini that $G(x_1)\colon=F_{x_1}$ is integrable and $$ \Big(\int G\,d\mu_1\Big)(x_2) =\int F(x_1,x_2)\,\mu_1(dx_1)~. $$ Moreover $$ \norm{f^n-G}_E=\norm{F^n-F}_{L_1(\mu)}, $$ and thus for $\mu_1$ almost all $x_1$: $f(x_1)=G(x_1)=F_{x_1}$.