Solution.

Continuity

\(f\) is bounded, say \(|f|\leq M\). For \(x<y\) in \([a,b]\), \[|F(y)-F(x)| = \left |\int _x^{y}f(t)\,dt\right | \leq M(y-x),\] so \(F\) is Lipschitz with constant \(M\), hence uniformly continuous. Continuity of \(F\) needs no continuity of \(f\) at all — only boundedness.

Differentiability where \(f\) is continuous

Let \(f\) be continuous at \(c\) and let \(\varepsilon >0\). Choose \(\delta >0\) with \(|f(t)-f(c)|<\varepsilon \) for \(|t-c|<\delta \). Then for \(0<|h|<\delta \), \[\left |\frac {F(c+h)-F(c)}{h} - f(c)\right | = \left |\frac {1}{h}\int _c^{c+h}[f(t)-f(c)]dt\right | \leq \frac {1}{|h|}\cdot |h|\varepsilon = \varepsilon .\] Hence \(F'(c) = f(c)\).

The two halves differ in what they need, and the difference is the content of the theorem: integration always smooths, producing a Lipschitz function from a merely bounded one, but it recovers the integrand as a derivative only where the integrand was already continuous.