Measure Theory 1

Let {X} be a nonempty set. A collection of subsets {\mathcal M\subset 2^X} is said to be a {\sigma}-algebra if

  • {\varnothing\in\mathcal M}
  • {E\in\mathcal M} implies {E^c\in\mathcal M}
  • If {\{E_n\}_{n=1}^\infty} is a countable sequence of sets in {\mathcal M}, then {\bigcup_{n=1}^\infty E_n\in\mathcal M}.

If {\mathcal E\subset 2^X}, the {\sigma}-algebra generated by {\mathcal E} is the intersection of all {\sigma}-algebras {\mathcal F} containing {\mathcal E}:

\displaystyle \sigma(\mathcal E) = \bigcap_{F\supset E}\mathcal F.

If {X} is a nonempty set and {\mathcal M} is a {\sigma}-algebra, the pair {(X,\mathcal M)} is called a measurable space. A function {\mu:\mathcal M\rightarrow[0,\infty]} is said to be a measure if the following conditions hold:

  • {\mu(\varnothing)=0}
  • If {\{E_n\}_{n=1}^\infty} is a countable sequence of disjoint sets in {\mathcal M}, then {\mu\left(\bigcup_{n=1}^\infty E_n \right)=\sum_{n=1}^\infty \mu(E_n)}.

If {(X,\mathcal M)} is a measurable space and {\mu} is a measure defined on {\mathcal M}, the triple {(X,\mathcal M,\mu)} is called a measure space.

Theorem 1 If {E,F\in\mathcal M} and {E\subset F}, then {\mu(E)\leqslant \mu(F)}.

Proof:

\displaystyle \mu(F) = \mu((F\cap E)\cup(F\cap E^c)) = \mu(F\cap E) + \mu(F\cap E^c) = \mu(E) + \mu(F\cap E^c)\geqslant \mu(E).

\Box

Theorem 2 If {E_1, E_2,\ldots\in\mathcal M} then

\displaystyle \mu\left(\bigcup_{n=1}^\infty E_n \right)\leqslant \sum_{n=1}^\infty \mu(E_n).

Proof: Put {F_1 = \varnothing} and {F_n = E_n\setminus E_{n-1}} for {n=2,3,\ldots}. Then for any {N}, {\bigcup_{n=1}^N F_n = \bigcup_{n=1}^N E_n = E_N}, and for any {i,j}, {F_i\cap F_j=\varnothing}. Hence

\displaystyle \mu\left(\bigcup_{n=1}^\infty E_n \right) = \mu\left(\bigcup_{n=1}^\infty F_n \right) = \sum_{n=1}^\infty \mu(F_n)\leqslant \sum_{n=1}^\infty \mu(E_n),

as {F_n\subset E_n}. \Box

Theorem 3 If {E_1, E_2, \ldots\in\mathcal M} and {E_n\subset E_{n+1}} for {n=1,2,\ldots} then

\displaystyle \mu\left(\bigcup_{n=1}^\infty E_n \right)=\lim_{n\rightarrow\infty}\mu(E_n).

Proof: Put {F_1 = \varnothing} and {F_n = E_n\setminus E_{n-1}} for {n=2,3,\ldots}. Then for any {N}, {\bigcup_{n=1}^N F_n = \bigcup_{n=1}^N E_n = E_N}, and for any {i,j}, {F_i\cap F_j=\varnothing}. Hence

\displaystyle  \begin{aligned} \mu\left(\bigcup_{n=1}^\infty E_n \right) &= \mu\left(\bigcup_{n=1}^\infty F_n \right)\\ &= \sum_{n=1}^\infty \mu(F_n)\\ &= \lim_{N\rightarrow\infty}\sum_{n=1}^N \mu(F_n)\\ &=\lim_{N\rightarrow\infty}\mu\left(\bigcup_{n=1}^N F_n \right)\\ &= \lim_{N\rightarrow\infty} \mu(E_N). \end{aligned}

\Box

Let {(X,\mathcal M,\mu)} be a measure space, and {(Y,\mathcal N)} be a measurable space. A function {f:X\rightarrow Y} is said to be measurable if for all {E\in\mathcal N}, {f^{-1}(E)\in\mathcal M}.

Theorem 4 If {\mathcal N=\sigma(\mathcal E)}, then {f:X\rightarrow Y} is measurable if and only if {f^{-1}(E)\in\mathcal M} for all {E\in\mathcal E}.

Proof: The set {\{E\subset \mathcal N: f^{-1}(E)\in\mathcal M \}} is a {\sigma}-algebra that contains {\mathcal M}. \Box

The Borel {\sigma}-algebra on {\mathbb R} is the {\sigma}-algebra generated by the open sets. If {E\in\mathcal M}, the characteristic function of {E} is the function {\chi_E: X\rightarrow\{0,1\}} defined by

\displaystyle \chi_E(x) = \begin{cases}1,& x\in E\\ 0,& x\notin E.\end{cases}

A simple function is a function of the form

\displaystyle  \varphi = \sum_{i=1}^n a_i\chi_{E_i},

where {a_i\geqslant 0} and {E_i\in\mathcal M}. For a simple function {\varphi}, define the Lebesgue integral by

\displaystyle \int_{\Omega}\varphi\,\mathsf d\mu = \sum_{i=1}^n a_i\mu(E_i).

Theorem 5 If {f:X\rightarrow[0,\infty)} is Borel-measurable, there exists a sequence of simple functions {\{\varphi_n\}} such that {\varphi_n\leqslant\varphi_{n+1}} and {\varphi_n} converges pointwise to {f}.

Proof: For each nonnegative integer {n} and each nonnegative integer {k<2^{2n}}, define

\displaystyle  \begin{aligned} E_{n,k} &= \left\{x\in X: \frac{k}{2^n}< f(x)\leqslant \frac{k+1}{2^n} \right\}\\ F_n &= \{x\in X: 2^n < f(x)\leqslant 2^{n+1} \}. \end{aligned}

Define

\displaystyle \varphi_n = \sum_{k=0}^{2^{2n}-1}\frac k{2^n}\chi_{E_{n,k}} + 2^n\chi_{F_n}.

Observe that for each {n},

\displaystyle \{x\in X: f(x)>0 \} = \left(\bigcup_{k=0}^{2^{2n}-1}E_{n,k}\right)\cup\left(\bigcup_{k=n}^\infty F_n \right),

and this union is disjoint. If {x\in E_{n,k}} then

\displaystyle \frac{k2^n}{2^{n+1}} < f(x)\leqslant \frac{(k+1)2^n}{2^{n+1}},

so that

\displaystyle \varphi_n(x) = \frac k{2^n} = \frac{2k}{2^{n+1}}\leqslant \varphi_{n+1}(x).

If {x\in F_n\cap F_{n+1}^c} then

\displaystyle \frac k{2^{n+1}}<f(x)\leqslant \frac{k+1}{2^{n+1}}

for some {2^{2n+1}\leqslant k <2^{2n+2}}, so that

\displaystyle \varphi_n(x) = 2^n = \frac{2^n2^{n+1}}{2^{n+1}}=\frac{2^{2n+1}}{2^{n+1}}\leqslant \frac k{2^{n+1}}=\varphi_{n+1}(x).

If {x\in F_n\cap F_{n+1}}, then

\displaystyle \varphi_n(x) = 2^n < 2^{n+1} = \varphi_{n+1}(x).

If {f(x)=\infty}, then {\lim_{n\rightarrow\infty}\phi_n(x)=\lim_{n\rightarrow\infty}2^n=\infty}. Otherwise, for sufficiently large {n}, {0\leqslant f(x)-\varphi_n(x)\leqslant\frac1{2^n}}, from which it follows that {\varphi_n\rightarrow f}. \Box

For {f:X\rightarrow[0,\infty]}, define the integral of {f} by the supremum of simple functions less than {f}:

\displaystyle \int_X f\,\mathsf d\mu = \sup_{\varphi\leqslant f}\int_X\varphi\,\mathsf d\mu.

For {f:X\rightarrow[-\infty,\infty]} define the positive and negative parts of {f} by

\displaystyle  \begin{aligned} f^+ &= \max\{f,0\}\\ f^- &= \max\{-f, 0\} \end{aligned}

Then {f=f^+-f^-} where {f^+\geqslant 0} and {f^-\geqslant0}, so we can define the integral of {f} by

\displaystyle \int_X f\,\mathsf d\mu = \int_X f^+\,\mathsf d\mu - \int_X f^-\,\mathsf d\mu.

For {f:X\rightarrow\mathbb C}, the integral is defined as

\displaystyle \int_X f\,\mathsf d\mu = \int_X (\mathrm{Re} f)^+\mathsf d\mu - \int_X(\mathrm{Re}f)^-\mathsf d\mu + \left(\int_X(\mathrm{Im} f)^+\mathsf d\mu - \int_X(\mathrm{Im} f)^-\mathsf d\mu \right)i.

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