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

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.

• That sounds pretty "formal" go me. You just have to note that a finite set has measure 0 so f= g almost everywhere and then use the criterion you cite. Nov 8, 2019 at 12:25

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 ...