Restriction of a finite measure to a set on unbounded function

Question

So I have a measure $(X,\mathscr{F},\mu)$, possibly finite, or $\sigma$-finite, or a completely general finite measure. $B\in \mathscr{F}$ is a set of finite measure. For every measure set $A\in \mathscr{F}$, define

$$\mu_B(A)=\mu(A\cap B)$$

I want to show that for every measurable function $f:X\to [0,\infty]$ (possibly unbounded,

$$\int f(x)\mu_B (dx)=\int f(x)1_B(x) \mu(dx)$$ Where $1_B(x)$ is the indicator function. The unbounded of the measure and the indicator threw me off. I am not sure how to start this one. This is a motivational observation in my textbook before the discussion of equality of interated integrals for $\sigma$-finite measure.

Thanks for any input.

Answer 1

This proof follows the "standard mantra".

1. Prove the case $f = 1_A$ where $A \subset X$ is measurable. (It's almost immediate from the definition.)

2. Prove the case where $f = \sum_{i=1}^n c_n 1_{A_n}$ is a simple function. (Use case 1 and linearity on both sides.)

3. Prove the general case. (Use the fact that every positive measurable function can be written as an increasing pointwise limit of simple functions, and use the monotone convergence theorem on both sides.)