# Lebesgue's fundamental theorem of calculus

The second part of the fundamental theorem of calculus states that

If $$f:[a,b]\to\mathbb{R}$$ is a Lebesgue-Integrable function and $$F$$ is a primitive of $$f$$, then $$\int_{a}^{b}f(x)dx=F(b)-F(a)$$.

I have been unable to prove this result by myself or find any reference for a proof of it.

# My attempt

Lets consider the function $$G(t):=f(a)+\int_{a}^{t}f(x)dx$$, which is absolutely continuous, so $$G$$ is differentiable almost everywhere, $$G'$$ is Lebesgue-Integrable and $$\int_{a}^{b}G'(x)dx=G(b)-G(a)=\int_{a}^{b}f(x)dx$$. So we have the desired conclusion for the function $$G$$ instead of a primitive $$F$$. We also know that $$G$$ is a primitive of $$f$$ almost everywhere, but I really don't know how to go further proving the equility for a primitive $$F$$.

• Rudin's RCA has a proof of this. – Kavi Rama Murthy Jun 1 at 8:22