I'm taking a course in measure theory and we defined integrability in a $\sigma$
-finite space as follows: Suppose $\left(X,\mathcal{F},\mu\right)$
is a $\sigma$-finite measure space, a measurable function $f:X\to\mathbb{R}$
is said to be integrable on $X$ (denoted $f\in L^{1}\left(X,\mathcal{F},\mu\right)$) if for every collection $\left\{ X_{m}\right\} _{m=1}^{\infty}$
such that $X_{m}\uparrow X$
, $X_{m}\in\mathcal{F}$
and $\mu\left(X_{m}\right)<\infty$
the following apply:
-
$f$
is integrable on every set $A\subseteq X$
such that $\mu\left(A\right)<\infty$
. -
The limit $\lim\limits _{m\to\infty}\int_{X_{m}}\left|f\right|d\mu$
exists and does not depend on the choice of $\left\{ X_{m}\right\} _{m=1}^{\infty}$
. -
The limit $\lim\limits _{m\to\infty}\int_{X_{m}}fd\mu$
does not depend on the choice of $\left\{ X_{m}\right\} _{m=1}^{\infty}$
.
If said conditions apply then we define $\int_{X}fd\mu=\lim\limits _{m\to\infty}\int_{X_{m}}\left|f\right|d\mu$
Now suppose $\mathcal{G}\subseteq\mathcal{F}$
is a $\sigma$
-algebra on $X$
. Let $f:X\to\mathbb{R}$
be a $\mathcal{G}$
-measurable function such that $f\in L^{1}\left(X,\mathcal{G},\mu\right)$
, is $f$
necessarily in $L^{1}\left(X,\mathcal{F},\mu\right)$
?
Obviously $\mathcal{G}$
-measurability implies $\mathcal{F}$
-measurability but what about integrability?
EDIT: It seems the construction of the integral we did is quite unorthodox, I'll elaborate further on the definitions: Suppose $\left(X,\mathcal{F},\mu\right)$ is a measure space and let $A\subseteq X$ be a subset of finite measure. We define a simple function $f:X\to\mathbb{R}$ to be any function taking a countable collection of real values $\left\{ y_{n}\right\} _{n=1}^{\infty}$. Denote $A_{n}=\left\{ x\in A\,|\, f\left(x\right)=y_{n}\right\}$. Assuming $f$ is measurable we say that $f$ is integrable on $A$ if the series ${\sum_{n=1}^{\infty}{\displaystyle y_{n}\mu\left(A_{n}\right)}}$ is absolutely convergent in which case we define: $$\int_{A}fd\mu={\displaystyle \sum_{n=1}^{\infty}}y_{n}\mu\left(A_{n}\right)$$
Furthermore, given any measurable function $f:X\to\mathbb{R}$ we say $f$ is integrable on $A$ if there is a sequence of simple functions (as defined) which are integrable on $A$ and converging uniformly to $f$ on $A$. In which case we define: $$\int_{A}fd\mu=\lim_{n\to\infty}\int_{A}f_{n}d\mu$$
Thanks in advance.
Best Answer
Okay, I will be more than happy to be corrected.
The limit must be unique for 2. The reason is that you may write the integral
$\int_{A_n} |f| = \int_X |f|1_{A_n}$. (1)
The integrand of the rhs is monotonically increasing, hence must have the same limit as putting the limit inside the integral by monotone convergence theorem. This is independent of the sets you choose as long they tend up to X so must be unique.
For 3, just use the fact $\int f = \int f^+ - \int f^-$ and apply 2.
I am somewhat unsatisfied with this argument. It feels wrong or i habve somehow cheated using Monotonce Convergence Theorem.