## Measures

Let $\left(X,ℳ\right)$ be a measurable space.

A positive measure on $\left(X,ℳ\right)$ is a function $\mu :ℳ\to \left[0,\infty \right]$ such that

if ${A}_{1},{A}_{2},\dots \in ℳ$ are such that if $i,j\in {ℤ}_{>0}$ and $i\ne j$ then ${A}_{i}\cap {A}_{j}=\varnothing$,
then $\mu \left(\bigcup _{i=1}^{\infty }{A}_{i}\right)=\sum _{i=1}^{\infty }\mu \left({A}_{i}\right)$.
A complex measure on $\left(X,ℳ\right)$ is a function $\mu :ℳ\to ℂ$ such that
if ${A}_{1},{A}_{2},\dots \in ℳ$ are such that if $i,j\in {ℤ}_{>0}$ and $i\ne j$ then ${A}_{i}\cap {A}_{j}=\varnothing$,
then $\mu \left(\bigcup _{i=1}^{\infty }{A}_{i}\right)=\sum _{i=1}^{\infty }\mu \left({A}_{i}\right)$.

A measure on $\left(X,ℳ\right)$ is a positive measure or a complex measure on $\left(X,ℳ\right)$.

## Integration with respect to positive measures

Let $\left(X,ℳ\right)$ be a measurable space. Let $\mu$ be a positive measure on $\left(X,ℳ\right)$ and let $E\in ℳ$. For a function $f:X\to \left[-\infty ,\infty \right]$ let

 ${f}^{+}=\frac{1}{2}\left(|f|+f\right)\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}{f}^{+}=\frac{1}{2}\left(|f|-f\right).$
For a simple measurable function
 $s=\sum _{i=1}^{n}{\alpha }_{i}{\chi }_{{A}_{i}}$      define      ${\int }_{E}s\phantom{\rule{0.1em}{0ex}}d\mu =\sum _{i=1}^{n}{\alpha }_{i}\mu \left({A}_{i}\cap E\right)$.
For a measurable function $f:X\to \left[0,\infty \right]$ define
 ${\int }_{E}f\phantom{\rule{0.1em}{0ex}}d\mu =\mathrm{sup}\left\{{\int }_{E}s\phantom{\rule{0.1em}{0ex}}d\mu \phantom{\rule{0.2em}{0ex}}|\phantom{\rule{0.2em}{0ex}}s\phantom{\rule{0.2em}{0ex}}\text{is simple measurable and}\phantom{\rule{0.2em}{0ex}}0\le s\le f\right\}$
For a measurable function $f:X\to \left[-\infty ,\infty \right]$ such that ${\int }_{E}{f}^{+}\phantom{\rule{0.1em}{0ex}}d\mu <\infty$ or ${\int }_{E}{f}^{-}\phantom{\rule{0.1em}{0ex}}d\mu <\infty$ define
 ${\int }_{E}f\phantom{\rule{0.1em}{0ex}}d\mu ={\int }_{E}{f}^{+}\phantom{\rule{0.1em}{0ex}}d\mu -{\int }_{E}{f}^{-}\phantom{\rule{0.1em}{0ex}}d\mu$.
For a measurable function $f:X\to ℂ$ let $f=u+iv$ where $u:X\to ℝ$ and $v:X\to ℝ$ are measurable and define
 ${\int }_{E}f\phantom{\rule{0.1em}{0ex}}d\mu ={\int }_{E}u\phantom{\rule{0.1em}{0ex}}d\mu +i{\int }_{E}v\phantom{\rule{0.1em}{0ex}}d\mu$.

## Integration with respect to complex measures

Let $\left(X,ℳ\right)$ be a measurable space. Let $\mu$ be a complex measure on $X$. Define a positive measure $|\mu |:ℳ\to \left[0,\infty \right]$ by

 $|\mu |\left(E\right)=\mathrm{sup}\left\{\sum _{i=1}^{\infty }|\mu \left({E}_{i}\right)|\phantom{\rule{1em}{0ex}}|\phantom{\rule{1em}{0ex}}{E}_{1},{E}_{2},\dots \in ℳ\phantom{\rule{0.5em}{0ex}}\text{partition}\phantom{\rule{0.5em}{0ex}}E\right\}$.
for a measurable function $f:X\to ℂ$ define
 ${\int }_{X}f\phantom{\rule{0.1em}{0ex}}d\mu ={\int }_{X}fh\phantom{\rule{0.1em}{0ex}}d|\mu |$,
where $h:X\to ℂ$ is a measurable function such that
 if $x\in X$ then $|h\left(x\right)|=1$,     and     if $E\in ℳ$ then $\mu \left(E\right)={\int }_{X}h\phantom{\rule{0.1em}{0ex}}d|\mu |$.

HW: Use the Radon-Nikodym theorem to show that the function $h$ exists (see [Ru, Theorem 6.12]).

## Notes and References

These notes were written for a course in "Measure Theory" at the Masters level at University of Melbourne. This presentation follows [Ru, Chapters 1-6].

## References

[Ru] W. Rudin, Real and complex analysis, Third edition, McGraw-Hill, 1987. MR0924157.