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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!

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@Srivatsan: Much obliged! – Tim Jan 14 '12 at 15:11
Why do you use the word "natural"? Notice that the Wikipedia quote you provided only claims that $\sigma$-finiteness is "convenient". Indeed, for, e.g. the Radon-Nikodym theorem, one way to prove it is to prove it for the finite case and then extend to the $\sigma$-finite case. – Quinn Culver Jan 14 '12 at 15:37
Egoroff's theorem works for a finite measured space, but not a $\sigma$-finite one. – Davide Giraudo Jan 14 '12 at 16:58
@DavideGiraudo: Thanks! What causes that? – Tim Jan 14 '12 at 17:01
I would say "the space is too big", for example taking $\mathbb R$ with the Borel $\sigma$-algebra and Lebesgue measure, taking $f_n(x)=\mathbf 1_{[n,n+1]}(x)$, it converges almost everywhere to $0$, but we can't find $A$ measurable such that $\mathbb R\setminus A$ has measure $\leq 1$, and the convergence is uniform on $A$. Indeed, $\mu(A)$ is infinite, so $\mu(A\cap [n,n+1])> 0$ for infinitely many $n$, and $\sup_{A}|f_n|\geq \sup_{A\cap [n,n+1]}|f_n|=1$. – Davide Giraudo Jan 14 '12 at 17:08
up vote 5 down vote accepted

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).

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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.

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Before making the proofs easier, they simply make the results hold true... Which might be your point. – Did Jan 14 '12 at 15:41

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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Thanks for expanding the comments into such a nice reply! – Tim Jan 21 '12 at 20:51

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