Localization to Finite-Measure Sets

Why Localize to Finite-Measure Sets?

An $L^p$ function can live on a measure space whose total measure is infinite. Working on finite-measure portions of that space is often useful, but it is not enough to note that those portions eventually cover the space: one must show that the omitted part of the function is small in $L^p$. Sigma-finiteness provides the appropriate portions, and integrability guarantees that the localization error tends to zero.

Throughout, $(X,\mathcal{M},\mu)$ is a measure space, not assumed complete or finite. Fix $p$ with $1\le p<\infty$. Functions may be real-valued or complex-valued and are measurable; $L^p(\mu)$ consists of functions, identified when equal $\mu$-a.e., for which $\int_X |f|^p\,d\mu<\infty$. The norm is $\|f\|_p=(\int_X|f|^p\,d\mu)^{1/p}$. For $E\in\mathcal{M}$, the restriction of $f$ to $E$, extended by zero on $E^c$, is written $f\chi_E$.

Definition: The measure space $(X,\mathcal{M},\mu)$ is sigma-finite if $X$ is a countable union of measurable sets of finite measure. An increasing finite-measure exhaustion is a sequence $(E_n)$ in $\mathcal{M}$ such that $E_n\subseteq E_{n+1}$, $\mu(E_n)<\infty$ for every $n$, and $\bigcup_{n=1}^{\infty}E_n=X$.

Building an Exhaustion

Sigma-finiteness does not mean that $X$ itself has finite measure. It means that $X$ can be covered by countably many finite-measure sets. Taking successive finite unions turns such a cover into an increasing exhaustion.

Proposition: Every sigma-finite measure space $(X,\mathcal{M},\mu)$ has an increasing finite-measure exhaustion. This assertion also covers the empty space.

Proof. If $X=\varnothing$, set $E_n=\varnothing$ for all $n$. Otherwise, sigma-finiteness gives measurable sets $A_j$ with $\mu(A_j)<\infty$ and $X=\bigcup_{j=1}^{\infty}A_j$. Define $E_n=\bigcup_{j=1}^{n}A_j$. Each $E_n$ is measurable, and $E_n\subseteq E_{n+1}$. Finite subadditivity gives $\mu(E_n)\le\sum_{j=1}^{n}\mu(A_j)<\infty$. Finally, $\bigcup_nE_n=\bigcup_jA_j=X$. Thus $(E_n)$ is the required exhaustion. $\square$

Conversely, the existence of such an exhaustion itself gives a countable cover by finite-measure sets, so it implies sigma-finiteness. An exhaustion is not unique. The localization result below holds for every increasing finite-measure exhaustion, not just one particular choice.

The Localization Theorem

Theorem: Let $(X,\mathcal{M},\mu)$ be a sigma-finite measure space, let $1\le p<\infty$, and let $f\in L^p(\mu)$ be real-valued or complex-valued. For every increasing finite-measure exhaustion $(E_n)$ of $X$, $$ \lim_{n\to\infty}\|f\chi_{E_n}-f\|_p=0. $$ The convergence is in $L^p$, meaning convergence in the norm $\|\cdot\|_p$; it does not assert uniform convergence.

The key calculation is that the error is exactly the part of $|f|^p$ on the omitted region. The Monotone Convergence Theorem says that if nonnegative measurable functions increase pointwise to a limit, their integrals increase to the integral of that limit. We apply it to the integrable function $|f|^p$ restricted to the exhaustion sets.

Proof. Choose a measurable representative of $f$. Since $f\in L^p(\mu)$, $\int_X|f|^p\,d\mu<\infty$, and $f$ is finite $\mu$-a.e.; changing its values on a measurable null set if necessary, take it finite everywhere. Define $g_n=|f|^p\chi_{E_n}$. Each $g_n$ is nonnegative and measurable. Since $E_n\subseteq E_{n+1}$, $g_n\le g_{n+1}$ at every point. Since $\bigcup_nE_n=X$, for every $x\in X$ there is an $N$ with $x\in E_N$, and then $g_n(x)=|f(x)|^p$ for every $n\ge N$. Thus $g_n$ increases pointwise to $|f|^p$. The Monotone Convergence Theorem therefore gives $\int_X g_n\,d\mu\to\int_X|f|^p\,d\mu$.

At each $x\in E_n$, $f(x)\chi_{E_n}(x)-f(x)=0$, and at each $x\in E_n^c$, $f(x)\chi_{E_n}(x)-f(x)=-f(x)$. Hence $$ \|f\chi_{E_n}-f\|_p^p =\int_X|f|^p\chi_{E_n^c}\,d\mu =\int_X|f|^p\,d\mu-\int_Xg_n\,d\mu. $$ The last equality follows by splitting the integral over the disjoint measurable sets $E_n$ and $E_n^c$; both terms are finite because $|f|^p$ is integrable. The first term on the right is fixed and finite, while the second tends to it by the preceding application of Monotone Convergence. The nonnegative difference therefore tends to zero. Taking its $p$-th root proves $\|f\chi_{E_n}-f\|_p\to0$. $\square$

This proof also gives an exact error formula, useful when the tail can be computed directly: $$ \|f\chi_{E_n}-f\|_p^p=\int_{E_n^c}|f|^p\,d\mu. $$ The formula applies to every measurable $E_n$, whether or not it has finite measure. The finite-measure condition matters because it makes each localized function supported on a finite-measure set; the error identity itself follows from the definition of the $L^p$ norm.

Worked Example: Exponential Decay on the Real Line

Take Lebesgue measure $m$ on $(\mathbb{R},\mathcal{L})$, let $p=2$, and set $f(x)=e^{-|x|}$. Use the exhaustion $E_n=[-n,n]$, $n\ge1$. Each set is measurable and $m(E_n)=2n<\infty$, and their union is $\mathbb{R}$. We calculate the squared localization error directly:

\|f\chi_{E_n}-f\|_2^2 =\int_{(-\infty,-n)\cup(n,\infty)}e^{-2|x|}\,dx =2\int_n^\infty e^{-2x}\,dx =2\left(\frac{e^{-2n}}{2}\right) =e^{-2n}.

Therefore $\|f\chi_{E_n}-f\|_2=e^{-n}$, which tends to zero. The integral over the whole line is $\int_{\mathbb{R}}e^{-2|x|}\,dx=2\int_0^\infty e^{-2x}\,dx=1$, so $f\in L^2(m)$. Check: the integral on $E_n$ is $2\int_0^n e^{-2x}\,dx=1-e^{-2n}$. Subtracting it from the total integral $1$ gives $e^{-2n}$, agreeing with the tail calculation.

Worked Example: A Function That Becomes Fully Included

Again use Lebesgue measure on $\mathbb{R}$, $1\le p<\infty$, and $E_n=[-n,n]$. Let $f=\chi_{(2,4)}$. This function is in $L^p(m)$ because $\int_{\mathbb{R}}|f|^p\,dm=m((2,4))=2$. For $n=1$, the support $(2,4)$ lies outside $E_1=[-1,1]$, so the error's $p$-th power is $2$. For $n=2$, the part of $(2,4)$ outside $[-2,2]$ is still $(2,4)$, of measure $2$. For $n=3$, the omitted part is $(3,4)$, of measure $1$. For $n\ge4$, every point of $(2,4)$ belongs to $[-n,n]$, so the omitted part is empty. Thus $$ \|f\chi_{E_n}-f\|_p= \begin{cases} 2^{1/p},& n=1\text{ or }n=2,\\ 1,& n=3,\\ 0,& n\ge4. \end{cases} $$ Endpoints do not change these integrals: singletons have Lebesgue measure zero, and the support here is the open interval $(2,4)$. Check: the error's $p$-th power is the measure of the omitted portion of $(2,4)$, since $|f|^p=1$ on that interval and $0$ elsewhere. Those measures are respectively $2,2,1,0$, as listed.

Worked Example: Geometric Sequence under Counting Measure

Let $X=\mathbb{N}=\{1,2,\ldots\}$ with counting measure, so $\int h\,d\mu=\sum_{k=1}^{\infty}h(k)$ for nonnegative $h$. Use $p=2$, $E_n=\{1,\ldots,n\}$, and $f(k)=2^{-k}$. Each $E_n$ has measure $n$, and their union is $\mathbb{N}$. The squared error is the tail of a geometric series:

\|f\chi_{E_n}-f\|_2^2 =\sum_{k=n+1}^{\infty}4^{-k} =\frac{4^{-(n+1)}}{1-\frac14} =\frac{4^{-n}}{3}.

Consequently, $\|f\chi_{E_n}-f\|_2=2^{-n}/\sqrt{3}\to0$. The function is in $L^2$, since $\sum_{k=1}^{\infty}4^{-k}=(1/4)/(1-1/4)=1/3$. Check: the squared norm of the restriction is $\sum_{k=1}^{n}4^{-k}=(1-4^{-n})/3$; subtracting this from the total squared norm $1/3$ gives $4^{-n}/3$, the same tail as above.

What the Hypothesis Prevents

Being on a sigma-finite space is not enough by itself. The theorem requires $f\in L^p(\mu)$. For a concrete failure, take Lebesgue measure on $\mathbb{R}$, $E_n=[-n,n]$, and $f(x)=1$. Although $(E_n)$ is an increasing finite-measure exhaustion, $f\notin L^p(m)$ for any finite $p$, because $\int_{\mathbb{R}}1\,dm=\infty$. More specifically, the proposed error has $$ \|f\chi_{E_n}-f\|_p^p=m(E_n^c)=\infty $$ for every $n$: the omitted set consists of two unbounded tails. A mistaken argument might say that the complements “shrink to nothing” and so their measures must tend to zero. In fact they decrease to the empty set, but each has infinite measure; continuity from above for measures does not give convergence to the measure of the intersection unless an appropriate finite-measure bound is available.

Do not confuse the measure of a region with the $L^p$ mass on it. The sets $E_n$ have finite measure, but their measures need not be bounded as $n$ grows. The theorem does not assert that $m(E_n)$ converges to a finite value. It asserts that the integral of $|f|^p$ over the omitted region tends to zero, which follows from integrability and the exhaustion.

Localization is useful because it separates an infinite-measure domain from the part of a function that matters in norm. The result applies to every finite-$p$ integrable function, including unbounded functions, and to any increasing finite-measure exhaustion. It makes no claim that the localized functions are bounded or that convergence is uniform. Its conclusion is precisely convergence in $L^p$.

Check Your Understanding

For each question, identify the measure space, the relevant exhaustion, and the quantity that measures the localization error.

Give an increasing finite-measure exhaustion of $\mathbb{R}$ with Lebesgue measure and verify the measure of each set.
For $f(x)=e^{-|x|}$ on $\mathbb{R}$, compute $\|f\chi_{[-n,n]}-f\|_1$.
Under counting measure on $\mathbb{N}$, let $f(k)=3^{-k}$ and $E_n=\{1,\ldots,n\}$. Compute the squared $L^2$ localization error.
Which step in the localization theorem uses the assumption that $f\in L^p(\mu)$?
Why does $E_n^c\downarrow\varnothing$ alone not imply $\mu(E_n^c)\to0$ on an infinite-measure space?

Localization to Finite-Measure Sets

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