Lebesgue integral of equal a.e. funtions

Question

Suppose that $f : \mathbb{R} \to \overline{\mathbb{R}}$ is a Lebesgue integrable function. Assume that $g = f$ a. e. Show that $g$ is also Lebesgue integrable and that $\int f = \int g$.

Could I have a hint?

Answer 1

$\int g=\int_{\{x|f(x)=g(x)\}}g(x)dx+\int_{\{x|f(x)\neq g(x)\}}g(x)dx=\int_{\{x|f(x)=g(x)\}}g(x)dx=\int_{\{x|f(x)=g(x)\}}f(x)dx=\int f$