Integral with respect to Restriction of Measure to Trace Sigma-Algebra of Measurable Set

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

Let $A \in \Sigma$.

Let $\Sigma_A$ be the trace $\sigma$-algebra of $A$ in $\Sigma$.

Let $\mu \restriction_{\Sigma_A}$ be the restriction of $\mu$ to $\Sigma_A$.

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

Let $f \restriction_A$ be the restriction of $f$ to $A$.

Define $f^\ast : X \to \overline \R$ by:


 * $\ds \map {f^\ast} x = \begin{cases}\map f x & x \in A \\ 0 & x \not \in A\end{cases}$

for each $x \in X$.

Then, we have:


 * $\ds \int \paren {f \restriction_A} \rd \paren {\mu \restriction_{\Sigma_A} } = \int f^\ast \rd \mu$

Proof
From Restriction of Measurable Function is Measurable on Trace Sigma-Algebra, $f \restriction_A$ is $\Sigma_A$-measurable.

Consider the case that $f$ is a positive simple function.

From Simple Function has Standard Representation, there exists:


 * a finite sequence $a_0, \ldots, a_n$ of non-negative real numbers
 * a partition $E_0, E_1, \ldots, E_n$ of $\Sigma_A$-measurable sets

such that:


 * $\ds f \restriction_A \, = \sum_{j \mathop = 0}^n a_j \chi_{E_j}$

From Trace Sigma-Algebra of Measurable Set:


 * $E_0, E_1, \ldots, E_n$ are in fact $\Sigma$-measurable subsets of $A$.

Then, since $\map {f^\ast} x = 0$ for $x \in X \setminus A$, we have:


 * $\ds f^\ast = \sum_{j \mathop = 0}^n a_j \chi_{E_j}$

Then $f^\ast$ is a positive simple function.

Then, we have:

Now suppose that $f$ is a general positive $\Sigma$-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 $x \in X$.

In particular:


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

for each $x \in A$.

We have established that $f^\ast$ (using the notation introduced in the hypotheses) is a positive simple function for each positive simple function $f$.

In particular, $f_n^\ast$ is a positive simple function for each $n \in \N$.

We now show that $f_n^\ast \to f^\ast$ pointwise.

Then we have, for $x \in A$:

For $x \in X \setminus A$, we have:

From Pointwise Limit of Measurable Functions is Measurable, we therefore have:


 * $f^\ast$ is $\Sigma$-measurable.

From our previous computation with positive simple functions, we have:


 * $\ds \int \paren {f_n \restriction_A} \rd \paren {\mu \restriction_{\Sigma_A} } = \int f_n^\ast \rd \mu$

Then, we have:

as required.