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Let $(\Omega, \mathcal{F}, \mu)$ be a measure space; consider a sub-$\sigma$-algebra $\mathcal{G}\subset \mathcal{F}$ and the restricted measure $\mu|_\mathcal{G} : A\in \mathcal{G} \rightarrow \mu(A) \in [0,+\infty]$, so that $(\Omega, \mathcal{G}, \mu|_\mathcal{G})$ is a measure space too.

Let $f: \Omega \rightarrow [0,+\infty[$ be a $\mathcal{G}$-measurable function, and consequently $f$ is a $\mathcal{F}$-measurable function too (obviously $[0,+\infty[$ is provided with the Borel $\sigma$-algebra).

In these hypothesis, for all $B\in \mathcal{G}$ we can consider the integrals:

$\int_B f d\mu$, done in the measure space $(\Omega, \mathcal{F}, \mu)$,

and

$\int_B f d\mu|_\mathcal{G}$, done in the measure space $(\Omega, \mathcal{G}, \mu|_\mathcal{G})$.

The question is: in general are these integrals equal? (for all $B\in \mathcal{G})$

According to the Lebesgue integral definition is:

$\int_B f d\mu = \sup\{$integrals of $\mathcal{F}$-measurable simple functions $s$ such that $\forall x\in B, s(x)\leq f(x)$ $\}$

and

$\int_B f d\mu|_\mathcal{G} = \sup\{$integrals of $\mathcal{G}$-measurable simple functions $s$ such that $\forall x\in B, s(x)\leq f(x)$ $\}$,

so my problem is: though the measure is the same on the subsets of $B$, having more measurable sets in $F$ implies that there could be more $\mathcal{F}$-measurable simple functions and the sup could be different; i'm not sure if the $\mathcal{G}$-measurability of $f$ is enough to guarantee the equality of the integrals.

Thanks for the help, i'm a beginner in measure theory.

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Notice that it suffices to establish this for $B=\Omega$. First, for a characteristic function of a $\mathcal{G}-$measurable set, the integrals clearly coincide; hence, by linearity, these integrals are the same for every $\mathcal{G}-$measurable simple function. Now apply the Monotone Convergence Theorem to conclude that they agree for every $\mathcal{G}-$measurable $f:\Omega \rightarrow [0,\infty]$.

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