Proof of $\int_a^b f(x) dx = \int_a^bg(x)dx$

Question

Let $f,g:[a,b] \to \mathbb{R}$ be functions, that coincide on all places except for a finite amounte of places.

How can one prove that if f is integrable then g is integrable as well and the following holds:

$$\int_a^b f(x) dx = \int_a^bg(x)dx$$

I would use the Lebesgue criterion which states that in a compact interval $[a,b]$ a bounded function is riemann-integrable if it's continuous almost everywhere in the interval. And if the function is riemann-integrable it also means that it is lebesgue integrable and that both integrals are identical. But I don't know how to do this formally.

Answer 1

Hint:

If $f = g$ except on a finite set $S$, then

$$\int_{[a,b]} (f-g) = \int_{[a,b]\setminus S} (f-g) + \int_{ S} (f-g)$$

The second integral on the RHS is $0$ because ...