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<p<\infty} 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

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