How do I prove that G is differentiable on (a,b)?

Are we talking Riemann Integrable? I'm going to assume so. Here is a sketch of the proof. I skipped some details in the paragraph below, so please let me know if there is anything you'd like fleshed out a bit more.

One of the requirements for a function f: J → ℝ with a compact domain J (such as [a; b]) to be Riemann integrable is that f be bounded on J -- otherwise, no matter how fine the mesh, there will always be partitions with arbitrarily large sums. So since f is bounded, its integral will be continuous. (In fact, given a bound M > 0 on f, the process of proving continuity actually ends up bearing out that F is K-Lipschitz continuous with K < M.)

Finally, we show that the integral G of the continuous function F is differentiable, in fact, G' = F. Let x ∈ J and ε > 0 be given. Since F is continuous at x, there exists δ > 0 such that for all z ∈ J with | z - x | < δ, we have | F(z) - F(x) | < ε. Hence

| F(x) - (G(z) - G(x)) / (z - x) |

= | F(x) - (∫ [ a,z ][ F(t)dt ] - ∫ [ a,x ][ F(t)dt ]) / (z - x) |

= | F(x) - ∫ [ z,x ][ F(t)dt ] / (z - x) |

= | ∫ [ z,x ][ F(x)dt ] / (z - x) - ∫ [ z,x ][ F(t)dt ] / (z - x) |

= | ∫ [ z,x ][ (F(x) - F(t))⋅dt ] / (z - x) |

≤ | ∫ [ z,x ][ | F(x)dt - F(t) |⋅dt ] / (z - x) |

≤ | ∫ [ z,x ][ ε⋅dt ] / (z - x) |

= | (z - x)⋅ε / (z - x) |

= ε.

Thus, since x and ε were arbitrary, we have shown that

[lim z→x ][ | F(x) - (G(z) - G(x)) / (z - x) | ] = 0,

or equivalently,

[lim z→x ][ (G(z) - G(x)) / (z - x) ] = F(x),

proving that G' = F.