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 $(\Omega, \mathcal{F}, \mu)$,


$\int_B f d\mu|_\mathcal{G}$, 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 $\forall x\in B, s(x)\leq f(x)$ $\}$


$\int_B f d\mu|_\mathcal{G} = \sup\{$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.


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 $f:\Omega \rightarrow [0,\infty]$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.