Suppose $f$ is Lebesgue integrable on $[a,b]$. Let $F(x)$=$\int_{a}^x fdt$. Then prove that $F$ is continuous on $[a,b]$.
I know that $F$ is continuous almost everywhere, because $F'(x)=f(x)$ almost everywhere on $[a,b]$. But does this imply that $F$ is continuous on $[a,b]$?
The fact that $F$ is differentiable almost everywhere on its own doesn't imply its continuity on $[a,b]$. Also you shouldn't be using that because it is a deeper fact (it uses the absolute continuity of $F$). The continuity of $F$ is more straightforward: suppose $(x_n)_n$ is a sequence tending to $x$ with $a\le x,x_n\le b$. Then,
$$|F(x)-F(x_n)| = \left|\int_y^{x} f(t) dt\right|\le \int_a^b \mathbf{1}_{[x_n,x]}(t) |f(t)| dt$$
The functions $g_n(t)=\mathbf{1}_{[x_n,x]}(t)|f(t)|$ converge to $0$ pointwise and are dominated by the Lebesgue integrable function $|f|$. Thus, by the dominated convergence theorem, the previous integral tends to $0$ as $n\to\infty$, which implies continuity of $F$ at $x$.
Nitpicky note: Here we mean $[x_n,x]$ to be the convex hull of the points $x$, $x_n$ regardless of which is larger.