If $\int_E f\,d\mu=\int_E g\,d\mu$ then $f=g$ a.e?

Question

Let $(E,\mathcal{A},\mu)$ be a finite measure space.

Let $f$ and $g$ two real-valued measurable functions such that $$\int_E f\,d\mu=\int_E g\,d\mu$$

Can we say that $f=g$ a.e $(*)$? Or it is necessary that: $\forall A\in \mathcal{A}$: such that $$\int_A f\,d\mu=\int_A g\,d\mu$$

I ask this question because I did not understand why in the book "Handbook of Multivalued Analysis Volume 1: Theory" written by "Shouchuan Hu", in Theorem 3.34 (page 187)

If $(*)$ not, what is the cause of why the author wrote this?

Answer 1

Definitely not.

$$\int_{-1}^1 \sin(x) dx = \int_{-1}^1 x dx = 0$$

The second assertion is true though, but you should say what you tried before we can help you.

Answer 2

Let $A\neq B$ be two sets st $\mu(A)=\mu(B)$, $f=I_A$ and $g=I_B$ their indicator functions. Then:

$$\int_E I_A\,d\mu=\int_E I_B\,d\mu \iff \mu(A)=\mu(B)$$

but clearly $f\neq g$.