Proving continuity for all $x$. I am having difficulty in proving the following problem.  Any hints would be greatly appreciated.
Let $f:[0,1]\times [0,1] \rightarrow \mathbb{R}$ be such that for each $y \in [0,1]$, $f(\cdot,y)$ is continuous and for each $ x \in [0,1]$, $f(x,\cdot)$ is measurable.  Assume that $f$ is also bounded.  Show that the function $F(x) = \int_{[0,1]}f(x,y)\,dy$ is continuous for all $x \in [0,1]$.

This is simply the Dominated Convergence Theorem.
Write $f_x(y) := f(x,y)$, then $F(x) = \int_0^1 f_x(y)dy$. For any $x_0 \in [0,1]$, as $x \to x_0$, $f_x \to f_{x_0}$ pointwise by continuity along the horizontal lines. Since $|f_x|$ is uniformly bounded for all $x$, we have by the DCT that
$$
\lim_{x \to x_0}F(x) = \lim_{x \to x_0}\int_0^1 f_x(y)dy = \int_0^1 \lim_{x \to x_0}f_x(y)dy = \int_0^1 f_{x_0}(y)dy = F(x_0)
$$
