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Let $\mathcal M_0$ be a $\sigma$-algebra that is contained in a $\sigma$-algebra $\mathcal M$ of subsets of a set $X$, $\mu$ a measure on $\mathcal M$ and $\mu_0$ the restriction of $\mu$ to $\mathcal M_0$. Let $f$ be a nonnegative real-valued function that is measurable with respect to $\mathcal M_0$ (and hence with respect to $\mathcal M$ as well). Then the set of all nonnegative simple functions $\phi\le f$ measurable with respect to $\mathcal M_0$ is a subset of the set of all nonnegative simple functions $\psi\le f$ measurable with respect to $\mathcal M$, and so $$ \int_Xf\,d\mu_0\le\int_Xf\,d\mu. $$ Can this inequality be strict?

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No, you have equality.

This is because you can find a nondecreasing sequence of nonnegative, $\mathcal M_0$-measurable functions $(\phi_n)$ such that $\phi_n(x)\to f(x)$ for all $x\in X$. By the monotone convergence theorem applied to $\mu$ and $\mu_0$ it follows that $\int fd\mu=\int fd\mu_0$.

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  • $\begingroup$ Thanks! This totally makes sense! $\endgroup$ – mathreader Jul 19 '13 at 12:09
  • $\begingroup$ If we change the question to say "Let $f$ be a real-valued function...", dropping the adjective "nonnegative", do the integrals still have to be equal? (I think the answer's yes, since you can probably just split $f$ up into positive and negative parts.) $\endgroup$ – xFioraMstr18 Jan 3 '18 at 23:20
  • $\begingroup$ Yes,the integrals remain equal, by splitting up indeed. $\endgroup$ – user370967 Mar 7 '19 at 13:47
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    $\begingroup$ The functions $\phi_n$ must also be simple. $\endgroup$ – user111 May 4 '19 at 19:08
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The intuition here is that integrating $\mathcal M_0$-measuarble function w.r.t. $\mu_0$ is actually integrating it w.r.t. $\mu$ since you are doing everything on the restriction where both measures agree. Let me provide a formal way of proving this: consider "two" measurable spaces $(X,\mathcal M)$ and $(X,\mathcal M_0)$ - this trick is often used in topology to study different typologies on the same set. You have that $$ \mathrm{id}:X\to X \quad\text{ given by }\quad \mathrm{id}(x) = x \;\forall x\in X \tag{1} $$ is a $\mathcal M/\mathcal M_0$-measurable map, and so induces the image measure $\mathrm{id}_*\mu$ on $(X,\mathcal M_0)$ define by $$ (\mathrm{id}_*\mu)(A) = \mu(\mathrm{id}^{-1}(A)) = \mu(A) \qquad\forall A\in \mathcal M_0. \tag{2} $$ From $(2)$ it is clear that we can represent $\mu_0 = \mathrm{id}_*\mu$, so that for any non-negative $f\in \mathcal M_0$: $$ \begin{align} \int f\,\mathrm d\mu_0 &= \int f\,\mathrm d(\mathrm{id}_*\mu) &\text{from }(2) \\ &= \int (f\circ \mathrm{id})\,\mathrm d\mu &\text{change of variables} \\ &= \int f\mathrm d\mu &\text{from }(1) \end{align} $$ where we used the change of variables formula.

Perhaps, there may be a more basic proof - but I believe it is important to learn techniques of dealing with different structures on the same set using the $\mathrm{id}$ map.

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  • $\begingroup$ Thank you! This is a little too advanced for me right now. But maybe I'll educate myself enough to understand it. $\endgroup$ – mathreader Jul 19 '13 at 12:11
  • $\begingroup$ @mathreader: sure :) for some reason from your questions and answers I got a feeling that such level won't be of a much problem to you, but anyways. I've edited to give some intuition of what we are aiming to prove first. $\endgroup$ – Ilya Jul 19 '13 at 12:16

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