Measure Theory 2

Time for some more measure theory. Given a measure space {(X,\mathcal M,\mu)}, let {L^+} denote the set of measurable nonnegative real-valued functions.

Theorem 1 \mbox{}

  1. Let {f,g\in L^+}. Then {\int (f+g)\mathsf d\mu = \int f \mathsf d\mu + \int g\mathsf d\mu}.
  2. Let {f\in L^+} and {c\in[0,\infty)}. Then {\int cf\mathsf d\mu = c\int f\mathsf d\mu}.
  3. Let {f,g\in L^+} with {f\leqslant g}. Then {\int f\mathsf d\mu \leqslant \int g\mathsf d\mu}.

Proof: \mbox{}

  1. If {\varphi = \sum_{i=1}^n a_i\chi_{E_i}} and {\psi = \sum_{j=1}^m b_j\chi_{F_j}} where we may assume WLOG

    \displaystyle \bigcup_{i=1}^n E_i = \bigcup_{j=1}^m F_j = X,


    \displaystyle \varphi = \sum_{i=1}^n a_i\chi_{E_i\cap\bigcup_{j=1}^m F_j}=\sum_{i=1}^n\sum_{j=1}^m a_i\chi_{E_i\cap F_j}

    and similarly

    \displaystyle  \psi = \sum_{j=1}^m\sum_{i=1}^n b_j \chi_{F_j\cap E_i}.


    \displaystyle \varphi + \psi = \sum_{i=1}^n\sum_{j=1}^m (a_i+b_j)\chi_{E_i\cap F_j},

    so that

    \displaystyle \begin{aligned} \int (\varphi+\psi)\mathsf d\mu &= \sum_{i=1}^n\sum_{j=1}^m (a_i+b_j)\mu(E_i\cap F_j) \\ &= \sum_{i=1}^n\sum_{j=1}^m a_i\mu(E_i\cap F_j) + \sum_{i=1}^n\sum_{j=1}^m b_j\mu(E_i\cap F_j)\\ &= \int \varphi \mathsf d\mu + \int\psi \mathsf d\mu. \end{aligned}

    From this it follows that if {\varphi_n\uparrow f} and {\psi_n\uparrow g}, then {(\varphi + \psi)_n\uparrow f+g} and hence by monotone convergence (to be proved later!)

    \displaystyle \begin{aligned} \int (f+g)\mathsf d\mu &= \int \lim_{n\rightarrow\infty} (\varphi_n + \psi_n)\mathsf d\mu\\ &= \lim_{n\rightarrow\infty} \int (\varphi_n + \psi_n)\mathsf d\mu\\ &= \lim_{n\rightarrow\infty} \int \varphi_n \mathsf d\mu + \int \psi_n\mathsf d\mu\\ &= \int f\mathsf d\mu + \int g\mathsf d\mu. \end{aligned}

  2. If {\varphi=\sum_{i=1}^n a_i\chi_{E_i}} then

    \displaystyle  \begin{aligned}\int c\varphi\mathsf d\mu &= \sum_{i=1}^n ca_i\mu(E_i)\\ &= c\sum_{i=1}^n a_i\mu(E_i)\\ &= c\int \varphi\mathsf d\mu.\end{aligned}

    Hence if {\varphi_n\uparrow f}, we have {c\varphi_n\uparrow cf}, and thus

    \displaystyle \int cf\mathsf d\mu = \lim_{n\rightarrow\infty} \int c\varphi_n\mathsf d\mu = \lim_{n\rightarrow\infty} c\int\varphi_n\mathsf d\mu = c\int f\mathsf d\mu.

  3. Since {g-f\geqslant 0}, this follows immediately from

    \displaystyle  0\leqslant \int (g-f)\mathsf d\mu = \int g\mathsf d\mu - \int f\mathsf d\mu.


One remark: For {f\in L^+} and {E\in\mathcal M} we define

\displaystyle \int_E f\mathsf d\mu = \int f\chi_E\mathsf d\mu.

It follows from monotonicity that if {E,F\in\mathcal M} with {E\subset F}, then for {f\in L^+}, {f\chi_E\leqslant f\chi_F} and hence

\displaystyle \int_E f\mathsf d\mu = \int f\chi_E\mathsf d\mu \leqslant \int f\chi_F\mathsf d\mu = \int_F f\mathsf d\mu.

Now for the monotone convergence theorem:

Theorem 2 Let {(X,\mathcal M,\mu)} be a measure space and {\{f_n\}} a sequence of measurable real-valued functions satisfying {f_n\geqslant 0}, {f_n\leqslant f_{n+1}} and {f_n\rightarrow f} pointwise. Then {f} is measurable and

\displaystyle \int f\mathsf d\mu = \lim_{n\rightarrow\infty}\int f_n\mathsf d\mu.

Proof: {\int f_n\mathsf d\mu} is a sequence of nondecreasing real numbers, and thus has a limit (possibly {+\infty}). Since {f_n\leqslant f} for each {n}, we have {\int f_n\mathsf d\mu\leqslant \int f\mathsf d\mu} and thus

\displaystyle \lim_{n\rightarrow\infty} \int f_n\mathsf d\mu \leqslant \int f\mathsf d\mu.

For the reverse inequality, let {\alpha\in(0,1)} and let {\varphi} be a simple function with {\varphi\leqslant f}. Set {E_n=\{x\in X: f_n(x)\geqslant\alpha\varphi(x) \}}. Since {f_n\rightarrow f} and {\alpha\varphi\leqslant f}, we have {\bigcup_{n=1}^\infty E_n = X}. Hence for each {n},

\displaystyle \int f\mathsf d\mu \geqslant \int_{E_n} f_n\mathsf d\mu \geqslant \alpha\int_{E_n}\varphi\mathsf d\mu.

Letting {n\rightarrow\infty} we obtain

\displaystyle \lim_{n\rightarrow\infty}\int f_n\mathsf d\mu \geqslant \alpha\int \varphi\mathsf d\mu.

Since this is true for all {0<\alpha<1}, it remains true for {\alpha=1}, and as {\varphi} was arbitrary, taking the supremum over {\varphi\leqslant f} we have

\displaystyle \lim_{n\rightarrow\infty}\int f_n\mathsf d\mu \geqslant \int f\mathsf d\mu.


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