Integral of Positive Measurable Function with respect to Restricted Measure

Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $\GG \subseteq \Sigma$ be a sub-$\sigma$-algebra of $\Sigma$.

Let $\mu \restriction_\GG$ be the restriction of $\mu$ to $\GG$.

Let $f : X \to \overline \R$ be a positive $\GG$-measurable function.

Then:


 * $\ds \int f \rd \mu = \int f \rd \mu \restriction_\GG$

Proof
We first prove the result for positive simple functions.

Let $f : X \to \R$ be a positive simple function that is $\GG$-measurable.

From Simple Function has Standard Representation there exists:


 * a finite sequence $a_1, \ldots, a_n$ of real numbers
 * a partition $E_1, \ldots, E_n$ of $\GG$-measurable sets

such that:


 * $f = \ds \sum_{j \mathop = 1}^n a_j \chi_{E_j}$

Then we have:

So we get the theorem for positive simple functions.

Now let $f$ be a general positive $\GG$-measurable function.

From Measurable Function is Pointwise Limit of Simple Functions, there exists an increasing sequence $\sequence {f_n}_{n \mathop \in \N}$ of positive simple functions such that:


 * $\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$

for each $x \in X$.

Then for each $n \in \N$ we have:


 * $\ds \int f_n \rd \mu = \int f_n \rd \mu \restriction_\GG$

From Integral of Positive Measurable Function as Limit of Integrals of Positive Simple Functions, taking $n \to \infty$ we obtain:


 * $\ds \int f \rd \mu = \int f \rd \mu \restriction_\GG$