# Measure Theory 3

Haven’t posted in two months! Time for some more measure theory, I suppose. Let’s start with Fatou’s lemma, which is a very useful tool:

Theorem 1 Let ${(X,\mathcal M,\mu)}$ be a measure space and ${f_n}$ a sequence of nonnegative measurable functions. Then

$\displaystyle \int_X\liminf_{n\rightarrow\infty}f_n\ \mathsf d\mu \leqslant \liminf_{n\rightarrow\infty} \int_X f_n\ \mathsf d\mu.$

Proof: Since ${\inf_{k\geqslant n}f_k\leqslant \inf_{k\geqslant n+1}f_k }$, this inequality is preserved by the integral and hence right-hand side is a monotone increasing sequence of real numbers, which has a limit in ${[0,\infty]}$. For any ${n}$, we have ${\inf_{k\geqslant n}f_k \leqslant f_k}$ for ${k\geqslant n}$, and hence

$\displaystyle \int_X \inf_{k\geqslant n}f_k\ \mathsf d\mu \leqslant \int_X f_k\ \mathsf d\mu \leqslant \inf_{k\geqslant n}\int_X f_k\ \mathsf d\mu,$

and by monotone convergence,

$\displaystyle \lim_{n\rightarrow\infty} \int_X \inf_{k\geqslant n}f_k\ \mathsf d\mu = \int_X \lim_{n\rightarrow\infty}\inf_{k\geqslant n} f_k\ \mathsf d\mu.$

Hence

$\displaystyle \int_X \lim_{n\rightarrow\infty}\inf_{k\geqslant n} f_k\ \mathsf d\mu = \int_X\liminf_{n\rightarrow\infty}f_n\ \mathsf d\mu \leqslant \lim_{n\rightarrow\infty}\inf_{k\geqslant n}\int_X f_k\ \mathsf d\mu = \liminf_{n\rightarrow\infty} \int_X f_n\ \mathsf d\mu.$

$\Box$

Note that this inequality requires no assumptions on ${f_n}$ (unlike monotone convergence). An immediate application is the so-called Dominated Convergence Theorem, but first, a digression. Given a measure space ${(X,\mathcal M,\mu)}$, we define

$\displaystyle L^1(\mu) := \left\{f:X\rightarrow\mathbb C \mid \int_X |f|\ \mathsf d\mu < \infty \right\}.$

From the definition of ${f = \Re(f)^+ - \Re(f)^-+\Im(f)^+-\Im(f)^-}$ it is clear that ${f\in L^1}$ if and only if the integrals of ${\Re(f)^+}$, ${\Re(f)^-}$, ${\Im(f)^+}$, and ${\Im(f)^-}$ are finite. For ${0 we define

$\displaystyle L^p(\mu) := \left\{ f:X\rightarrow\mathbb C \mid f^p\in L^1(\mu) \right\},$

i.e. ${f\in L^p}$ if and only if

$\displaystyle \int_X |f|^p\ \mathsf d\mu < \infty.$

Now, we would like to define a norm on ${L^p}$ by e.g.

$\displaystyle \|f\|_p = \left(\int_X |f|^p\right)^{\frac1p}.$

The problem is that this is only a pseudonorm, as ${\|0\|_p=0}$, but also ${\|f\|_p=0}$ for ${f=0}$ a.e. To remedy this, we consider the equivalence relation ${\sim}$ on ${L^p}$ where

$\displaystyle f\sim g \iff f = g\ \mathrm{ a.e.}$

Then we define

$\displaystyle \mathcal L^p = L^p/\sim$

to be the set of equivalence classes of ${L^p}$ under ${\sim}$. That is, we consider functions that agree except on a set of measure zero (which does not change the integral) to be equivalent. Then ${\|\cdot\|_p}$ defines a norm on ${\mathcal L^p}$, and ${\rho(f,g) = \|f-g\|_p}$ a metric. We will discuss ${\mathcal L^p}$ spaces in more detail later. Back to the Dominated Convergence Theorem:

Theorem 2 Let ${f_n}$ be a sequence of measurable functions with ${f_n\rightarrow f}$ pointwise and suppose there exists a function ${g\in L^1}$ such that ${|f_n|\leqslant g}$ for all ${n}$. Then ${f\in L^1}$ and

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

Proof: Since ${|f_n|\leqslant g}$ for all ${n}$, letting ${n\rightarrow\infty}$ we have ${|f|\leqslant g}$, so

$\displaystyle \int_X |f|\ \mathsf d\mu \leqslant \int_X g\ \mathsf d\mu < \infty,$

and hence ${f\in L^1}$. Now by Fatou’s lemma,

$\displaystyle \int_X \liminf_{n\rightarrow\infty}(g + f_n)\ \mathsf d\mu \leqslant \liminf_{n\rightarrow\infty}\int_X (g+f_n)\ \mathsf d\mu = \int_X g\ \mathsf d\mu +\liminf_{n\rightarrow\infty} \int_X f_n \ \mathsf d\mu$

and

$\displaystyle \int_X \liminf_{n\rightarrow\infty}(g-f_n) \ \mathsf d\mu \leqslant \liminf_{n\rightarrow\infty} \int_X (g-f_n)\ \mathsf d\mu = \int_X g\mathsf d\mu - \limsup_{n\rightarrow\infty} \int_X f_n\mathsf d\mu.$

Since ${\int_X g\ \mathsf d\mu <\infty}$ we conclude that

$\displaystyle \limsup_{n\rightarrow\infty} \int_X f_n\ \mathsf d\mu \leqslant \liminf_{n\rightarrow\infty} \int_X f_n \ \mathsf d\mu \left(= \lim_{n\rightarrow\infty} \int_X f_n \ \mathsf d\mu \right),$

and hence by another application of Fatou’s lemma,

$\displaystyle \int_X f\ \mathsf d\mu \leqslant \liminf_{n\rightarrow\infty} \int_X f_n \ \mathsf d\mu=\lim_{n\rightarrow\infty} \int_X f_n \ \mathsf d\mu \leqslant \int_X \limsup_{n\rightarrow\infty} f_n \ \mathsf d\mu = \int_X f\mathsf d\mu.$

$\Box$