Limit of integral - part 2 Inspired by the recent post "Limit of integral", I propose the following problem (hoping it will not turn out to be too easy). Suppose that $g:[0,1] \times [0,1] \to {\bf R}$ is continuous in both variables separately. Is it true that, for all $x_0 \in [0,1]$,
$$
 \lim \limits_{x \to x_0 } \int_0^1 {g(x,y)\,{\rm d}y}  = \int_0^1 {g(x_0 ,y)\,{\rm d}y} .
$$ 

I think I have another counterexample.  Define $f(x)=\int_{0}^x e^{-1/t}dt$ for $x\gt0$.  This is chosen because it goes to zero as $x$ goes to zero from the right and because experimentation led me to the differential equation $\frac{f''(x)}{f'(x)}=\frac{1}{x^2}$ as a sufficient condition for the following to work.  Define 
$$g(x,y) = \left\{
     \begin{array}{lr}
       \frac{xy}{f(x)+y^2} & \text{if } x\gt0, \\
       0 & \text{if } x=0, 
     \end{array}
   \right.$$
for $(x,y)$ in $[0,1]\times[0,1]$, and let $x_0=0$.  The right hand side of your tentative equation is obviously $0$.  The left hand side is 
$$\lim_{x\to0+}\frac{x}{2}\log\left(1+\frac{1}{f(x)}\right),$$
which comes out to $\frac{1}{2}$ after $2$ applications of l'Hôpital's rule, if I did it correctly.
(I started by playing with the standard example of a discontinuous but separately continuous function on $\mathbb{R}^2$, $f(x,y)=\frac{xy}{x^2+y^2}$ when $x$ or $y$ is nonzero, $f(x,y)=0$ when $x=y=0$. Then I tried to see how the $x^2$ in the denominator could be modified to give a counterexample here, by replacing it with an unknown $f(x)$ that goes to $0$ at $0$ and seeing what further properties of $f(x)$ would make it work.  As I mentioned, this led in particular to the sufficient condition $\frac{f''(x)}{f'(x)}=\frac{1}{x^2}$, and hence to this example.  Unfortunately, I can't offer any real intuition.)
Added
I decided to look at this a little more, and came up with an example simpler than the other one I gave.  With $$g(x,y)=\frac{-y}{\log(x/2)(x+y^2)}$$ 
for $x\gt0$ and $0$ otherwise, the same result as above holds.
