5-6

We let (h_n) be family such that \lim_{n\to \infty}h_n=0 and for any x\geq 0, define 
f_n(y)=\frac{f(x+h_n,y)-f(x,y)}{h_n} 
Since\int f(x,y)dy exists for all x\geq 0, f(x,y) is measurable function of y 
Also, addition of measurable functions and scalar multiplication of measurable function is measurable, f_n(y) are measurable for all n 
Now, for any x\geq 0 
\lim_{n\to \infty}f_n(y)=\lim_{n\to\infty}\frac{f(x+h_n,y)-f(x,y)}{h_n}=
\lim_{h\to 0}\frac{f(x+h,y)-f(x,y)}{h}=\frac{\partial f}{\partial x}(x,y) 
The family(f_n) is pointwise convergence to \frac{\partial f}{\partial x}(x,y) hence \frac{\partial f}{\partial x}(x,y) is measurable on y. This proof the first claim. 
For the second assertion, for any \epsilon>0, there is some N>0 such that for all n\geq N, we have 
|f_n(y)|-|\frac{\partial f}{\partial x}(x,y)|\leq |f_n(y)-\frac{\partial f}{\partial x}(x,y)|<\epsilon\implies 
|f_n(y)|<\epsilon+|\frac{\partial f}{\partial x}(x,y)|<\epsilon +M, Mcomes from the hypothesis that \frac{\partial f}{\partial x}(x,y) is bounded. Since constants are integrable, by Lebesgue dominated convergence theorem, 
\lim_{n\geq N,n\to\infty}\int f_n(y)dy=\int{\frac{\partial f}{\partial x}}(x,y)dy. But 
\lim_{n\geq N,n\to\infty}\int f_n(y)dy=
\lim_{n\geq N,n\to\infty}\frac{1}{h_n}\int f(x+h_n,y)-f(x,y)dy=
 
\lim_{h\to 0}\frac{1}{h}\int f(x+h,y)-f(x,y)dy=
\lim_{h\to 0}\frac{\int f(x+h,y)dy-\int f(x,y)dy}{h}=
 
\frac{d}{dx}\int f(x,y)dy
, The equality follows.