Significance of $\sigma$-finite measures

Question

From Wikipedia:

The class of $\sigma$-finite measures has some very convenient properties; $\sigma$-finiteness can be compared in this respect to separability of topological spaces. Some theorems in analysis require σ-finiteness as a hypothesis. For example, both the Radon–Nikodym theorem and Fubini's theorem are invalid without an assumption of $\sigma$-finiteness (or something similar) on the measures involved.

Though measures which are not σ-finite are sometimes regarded as pathological, ...

I was wondering what makes $\sigma$-finite measures so natural to mathematicians (they often think of them in the first place when it comes to measures, while I as a layman don't have that instinct), well-behaved (as opposite to "pathological") and important (appearing in conditions in many theorems such as Radon-Nikodym, Lebesgue decomposition and Fubini's Theorems)?

In what sense/respect, can $\sigma$-finiteness be compared to separability of topological spaces?

For example, are most or all properties true for finite measures also true for $\sigma$-finite measures, but not for general infinite measures? If yes, why is that?

Are all above because of equivalence of $\sigma$-finite measures to probability measures? If yes, how is it the reason?

Thanks and regards!

Answer 1

When it comes to intuition, $\sigma$-finiteness is a property that can be said to be part of the intuition for measures in the first place. That a set has "infinite size" means intuitively that it consists of lots of parts of small size, not of few parts of "infinite size". There exists other classes of measure spaces that follow this intuition to some degree (strictly localizable measure spaces), but $\sigma$-finiteness is the most natural one.

Now let $(X,\Sigma,\mu)$ be a $\sigma$-finite measure space. Then there exists a countable family of finite measure spaces $(X_i,\Sigma_i,\mu_i)$ such that

1. the $X_i$ partition $X$,
2. a set $A$ is in $\Sigma$ if and only if $A\cap X_i\in\Sigma_i$ for all $i$,
3. and then $\mu(A)=\sum_i\mu_i(A\cap X_i)$.

So one can think about $\sigma$-finite measure spaces as a family of finite measure spaces lying "side by side". That there are only countably many of them makes sure that combining them works well. This decomposition makes it for example obvious that the Radon-Nikodym theorem for $\sigma$-finite measure spaces is really no more general than the Radon-Nikodym theorem for finite measure spaces.

Finally: I don't think one should make too much of an anology with separable topological spaces, there are conditions in measure theory that can more naturally be seen as an analog to separability (being countably generated and having countable Maharam type).

Answer 2

I will focus my answer on the properties which are true for the finite measure spaces but not $\sigma$-finite ones. Recall Egoroff's theorem:

Let $(X,\mathcal A,\mu)$ a finite measured space, and $\{f_n\}$ a sequence of measurable function from $X$ to $\mathbb R$ endowed with the Borel $\sigma$-algebra. If $f_n\to 0$ almost everywhere then for each $\varepsilon>0$ we can find $A_{\varepsilon}\in\mathcal A$ such that $\mu(X\setminus A_{\varepsilon})\leq\varepsilon$ and $\sup_{x\in A_{\varepsilon}}|f_n(x)|\to 0$.

It's not true anymore if $(X,\mathcal A,\mu)$ is not finite. For example, if $X=\mathbb R$, $\mathcal A=\mathcal B(\mathbb R)$ and $\mu=\lambda$ is the Lebesgue measure, taking $f_n(x)=\begin{cases}1&\mbox{ if }n\leq x\leq n+1,\\\ 0&\mbox{otherwise}, \end{cases}$ we can see that $f_n\to 0$ almost everywhere, but if $A$ is such that $\lambda(\mathbb R\setminus A)\leq 1$, then $\mu(A)=+\infty$, hence $A\cap [n,n+1]$ has a positive measure for infinitely many $n$, say $n=n_k$, so $\sup_A|f_{n_k}|\geq \sup_{A\cap [n_k,n_k+1]}|f_{n_k}|=1$.

An explanation could be the following: if $(X,\mathcal A,\mu)$ is $\sigma$-finite, $\{A_n\}$ is a partition of $X$ into finite measure sets, and a sequence converges almost everywhere on $X$, then we have the convergence in measure on each $A_n$: for $k$ and for a fixed $\varepsilon>0$ we can find a $N(\varepsilon,k)\in\mathbb N$ such that $\mu(\{|f_n|\geq \varepsilon\}\cap A_k)\leq \varepsilon$ if $n\geq N(\varepsilon,k)$. The problem, as the counter-example show, is that this $N$ cannot be chosen independently of $k$.

An other result:

Let $(X,\mathcal A,\mu)$ a finite measured space, and $\{f_n\}$ a sequence which converges almost everywhere to $0$. Then $f_n\to 0$ in measure.

We can use the same counter-example as above.

Inclusions between $L^p$ space may change whether the measured space is finite. If $(X,\mathcal A,\mu)$ is a finite measured space, then for $1\leq p\leq q\leq \infty$ we have $L^q(X,\mathcal A,\mu)\subset L^p(X,\mathcal A,\mu)$, as Hölder's inequality shows. But with $X=\mathbb N$, $\mathcal A=2^{\mathbb N}$ and $\mu$ the counting measure, we have for $1\leq p\leq q\leq \infty$, $\ell^p\subset l^q$, so the inclusions are reversed.

Answer 3

Well, first of all, Lebesgue measure on $\mathbb{R}^n$ and counting measure on $\mathbb{Z}^n$ are both $\sigma$-finite, so at least it is a familiar setting. I would say that $\sigma$-finiteness is nice for the same reason that all finiteness conditions are nice: they make proofs easier, and reasonable spaces satisfy them. A $\sigma$-finite measure space is one that can be exhausted by a sequence of finite measure spaces, which makes them "close enough" to finite measure spaces that many proofs go through in this setting that don't go through in a more general setting.

$\sigma$-finiteness is really more closely analogous to $\sigma$-compactness than to separability, but both conditions assert that a space can be "analyzed countably" in some sense.

Answer 4

Adding to all the other responses here, $\sigma$ finiteness is natural because if $f$ is integrable, then the set: $$ \{|f| > 0\} $$ is $\sigma$ finite (countable union of sets of finite measure) in any measure space. Indeed, this comes from Markov's inequality: $$ \{|f| > 0\} = \bigcup_{n=1}^\infty \left\{|f| > \frac{1}{n}\right\} $$ For each $n \in \mathbb{N}$ we get: $$ \mu\left(\left\{|f| > \frac{1}{n}\right\}\right) \leq n\|f\|_{L^1} < \infty $$ Of course the size of the sets will be increasing, but they are all finite. Thus, on any measure space, once you have an integrable function on that space you are naturally in a $\sigma$ finite world. In general a measure space is sigma finite when it admits a strictly positive, integrable real valued function.

Answer 5

-finiteness can be compared in this respect to separability of topological spaces

In a separable topological space a countable dense set can be used to approximate any element of the topological space, in the sense that any environment around any element contains a member of the countable dense set. This also allows you to construct, in a metric space, a sequence of the countable dense set which converges to your desired element.

In the same sense, you can construct using a "sigma-finite set" of measurable sets $E_n$ a "measure-approximation" of $A$ by utilizing cuts of $E_n$ with respect to $A$. We will use below the two most common characterizations for $\sigma$-finiteness, that of a countable increasing monotone sets and that of the countable partitioning sets (countable pairwise disjoint cover):

let $(\Omega,\mathcal{A},\mu)$ be a $\sigma$-finte measure space and let $A\in\mathcal{A}$, then:

$$ \begin{align*} &E_n\uparrow\Omega &\Rightarrow \mu(A) = \lim_{n\to\infty} \mu(E_n\cap A) \\ &\Sigma_{n=1}^{\infty} E_n = \Omega &\Rightarrow \mu(A)=\Sigma_{n=1}^{\infty} \mu(E_n\cap A) \end{align*} $$

And then just like in topology, this sequence of "sigma-finite se cuts of a set" ($E_n\cap A$) allow you to approximate measure-theoretic properties of the set, where the measures of $E_n$ work as determining mediators of measure-wise properties. This is the significance of $\sigma-$finite measures.

I believe the main examples given of Fubini and Radon–Nikodym are too complicated as examples for this nice, fundamental property. Look at the proof of the uniqueness theorem for measures which are are $\sigma-$finite on a $\cap-$closed generators of the $\sigma-$Algebra. How do we use the $\sigma-$finiteness of the two measures to mediate their measure-theoretic property (e.g. their equality on the generator elements) to the entire $\sigma-$Algebra?