In reading my measure theory book I found the following interesting problem, which I honestly have no idea how to even attempt:
Suppose $\mathcal{S}$ and $\mathcal{T}$ are $\sigma$ -algebras on a set $X$ and $\mathcal{S} \subset \mathcal{T} .$ Suppose $\mu_{1}$ is a measure on $(X, \mathcal{S}), \mu_{2}$ is a measure on $(X, \mathcal{T}),$ and $\mu_{1}(E)=\mu_{2}(E)$ for all $E \in \mathcal{S} .$ Prove that if $f: X \rightarrow[0, \infty]$ is $\mathcal{S}$ -measurable, then $\int f d \mu_{1}=\int f d \mu_{2}$
Does anyone have any suggestions how to go about this proof?