Measure Theory – Significance of $\sigma$-Finite Measures

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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!

Best Answer

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.

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