Theorem. Let $(X,\mathcal B,\mu)$ a finite measure space, where $\mu$ is a positive measure. Let $\mathcal A\subset \mathcal B$ an algebra generating $\cal B$.

Then for all $B\in\cal B$ and $\varepsilon>0$, we can find $A\in\cal A$ such that $$\mu(A\Delta B)=\mu(A\cup B)-\mu(A\cap B)<\varepsilon.$$

I don't think there is a proof in this site.

It's a useful result for several reasons:

  • We know what the algebra generated by a collection of sets is, but not what the generated $\sigma$-algebra is.
  • The map $\rho\colon \cal B\times\cal B\to \Bbb R_+$, $\rho(A,A')=\mu(A\Delta A')$ gives a pseudo-metric on $\cal B$. This makes a link between generating for an algebra and dense for the pseudo-metric.
  • We say that a $\sigma$-algebra is separable if it's generated by a countable class of sets. In this case, the algebra generated by this class is countable. An with the mentioned result, we can show that $L^p(\mu)$ is separable for $1\leq p<\infty$, which makes a link between the two notions.
  • In ergodic theory, we have to test mixing conditions only an a generating algebra, not on all the $\sigma$-algebra.
  • $\begingroup$ In the case of a countably generated sigma algebra, I believe that one can also give a proof using the martingale convergence theorem (assuming that one has first covered martingales). $\endgroup$
    – shalop
    Aug 18, 2017 at 5:24
  • $\begingroup$ @Shalop Could you sketch how that would go? $\endgroup$
    – aduh
    Aug 9, 2018 at 13:26
  • 1
    $\begingroup$ @aduh Let $\mathcal F = \sigma(E_i:i\in\Bbb N)$ be countably generated. For each $n$ let $\mathcal F_n:= \sigma(E_i:i\leq n)$ so that $\mathcal F_n$ form a filtration. Now take a set $A \in \cal F$ and consider the martingale $X_n:=\Bbb E[1_A|\mathcal F_n]$. This converges a.s. to $1_A$, from which one may conclude easily enough (just to clarify, this would only prove that $\mathcal F$ can be approximated arbitrary closely by elements of the algebra which is generated by the $E_i$)... $\endgroup$
    – shalop
    Aug 9, 2018 at 16:24

3 Answers 3


Proof: Let $$\mathcal S:=\left\{A\in \mathcal{B}\mid \forall\varepsilon>0,\exists A'\in\mathcal A,\mu(A\Delta A')\leq \varepsilon\right\}.$$ We have to prove that $\cal S$ is a $\sigma$-algebra, as it contains by definition $\cal A$.

  • $X\in\cal S$ since $X\in\cal A$.
  • If $A\in\cal S$ and $\varepsilon>0$, let $A'\in\cal A$ such that $\mu(A\Delta A')\leq \varepsilon$. Then $\mu(A^c\Delta A'^c)=\mu(A\Delta A')\leq \varepsilon$ and $A'^c\in\cal A$.
  • First, we show that $\cal A$ is stable by finite unions. By induction, it is enough to do it for two elements. Let $A_1,A_2\in\cal S$ and $\varepsilon>0$. We can find $A'_1,A'_2\in\cal A$ such that $\mu(A_j\Delta A'_j)\leq \varepsilon/2$. As $$(A_1\cup A_2)\Delta (A'_1\cup A'_2)\subset (A_1\Delta A'_1)\cup (A_2\Delta A'_2),$$ and $A'_1\cup A'_2\in\cal A$, $A_1\cup A_2\in \cal A$.

    Now, let $\{A_k\}\subset\cal S$ pairwise disjoint and $\varepsilon>0$. For each $k$, let $A'_k\in\cal A$ such that $\mu(A_k\Delta A'_k)\leq \varepsilon 2^{-k}$.
    Let $N$ be such that $\mu\left(\bigcup_{j\geq N+1}A_j\right)\leq \varepsilon/2$ (such a choice is possible since $\bigcup_{j\geqslant 1}A_j$ has a finite measure and $\mu\left(\bigcup_{j\geq n+1}A_j\right)\leq \sum_{j\geq n+1}\mu\left(A_j\right)$ and this can be made arbitrarily small). Let $A':=\bigcup_{j=1}^NA'_j\in\cal A$. As $$\left(\bigcup_{k\geq 1}A_k\right)\Delta A'\subset \bigcup_{j=1}^N(A_j\Delta A'_j)\cup\bigcup_{k\geq N+1}A_k,$$ and we conclude by sub-additivity.

  • $\begingroup$ If you put $\mathcal{S}' = \{A \in \mathcal{A}|\,\mu(A) = \infty\} \cup \mathcal{S}$, then $\mathcal{S}'$ is a $\sigma$-algebra even when the measure is not finite. Then you get that either $\mu(A) = \infty$ or $A$ can be "approximated" by some set in $\mathcal{A}$. Right? $\endgroup$ Sep 19, 2013 at 16:08
  • $\begingroup$ Sorry. I think I missed the "complement" part $A^c \in \mathcal{S}'$, so my comment above is probably not right. $\endgroup$ Sep 19, 2013 at 16:18
  • 2
    $\begingroup$ FWIW, for future references it seems to be Exercise 4.7.63 of the first volume. $\endgroup$
    – SBF
    Jul 11, 2014 at 11:46
  • 1
    $\begingroup$ @AndréCaldas to show that $\mathcal{S}$ is a sigma-algebra, we need to show that if $\{A_k\} \subset \mathcal{S}$ where $\{A_k\}$ are not necessarily disjoint, then $\cup_k A_k \in \mathcal{S}$. Why did you assume they are disjoint? $\endgroup$
    – m0_as
    Feb 17, 2016 at 1:04
  • 3
    $\begingroup$ This result is also given in the book Measure Theory by Halmos (Page 56, Theorem D, Section 13) $\endgroup$
    – jpv
    Jun 7, 2017 at 6:34

I have found in Probability And Examples, by Rick Durret, second edition, 1996, the following question as an exercise: (Page 452, Appendix: Measure theory, Exercise 3.1):

Let $\mathcal{A}$ be an algebra, $\mu$ a measure on $\sigma(\mathcal{A})$ and $B \in \sigma(\mathcal{A})$ with $\mu(B) < \infty$ . Then for any $\epsilon > 0$ there is an $A \in \mathcal{A}$ with $\mu(A\Delta B) < \epsilon$.

The interesting fact is that there is no any assumption here that the measure is finite or even $\sigma$ finite! The ONLY assumption is that the measure of $B$ is finite. Another interesting fact is that this question was omitted in the 4-th edition of the book. (I didn't find it...)

So is it a mistake?

If the measure is not finite, i.e. $\mu(\Omega) = \infty$, then the collection of all well approximated sets may not be a $\sigma$ algebra! The regular proof fails when one has to show that it is closed under countable unions. It is still an algebra, and it is also closed under countable unions which have finite measure. But it may (?) not be closed under general countable unions...

However, if the restriction of the measure $\mu|_{\mathcal{A}}$ is $\sigma$ finite on $\mathcal{A}$ then this restriction has a unique extension to the $\sigma(\mathcal{A})$, i.e. $\mu$ defined on $\sigma(\mathcal{A})$ must coincide with $(\mu|_{\mathcal{A}})^*$ (the outer measure obtained by $\mu|_{\mathcal{A}}$)

Hence: $(\mu|_{\mathcal{A}})^*(B) < \infty$ which is equivalent to: $$\mu(B) = (\mu|_{\mathcal{A}})^*(B) = \inf{\left\{\sum_{i=1}^{\infty}{\mu(A_i)} : B \subseteq \bigcup_{i = 1}^{\infty}{A_i}, A_i \in \mathcal{A}\right\}} < \infty$$

So I can pick up some countable covering of $B$ (with disjoint sets from $\mathcal{A}$) for which: $$\mu(B) \le \sum_{i=1}^{\infty}{\mu(A_i)} \le \mu(B) + \epsilon/2 $$ and then drop the tail of the series and get some finite subcovering which will approximate $B$.

So under the additional assumption that the restriction $\mu|_\mathcal{A}$ is $\sigma$ finite, this result is true.

But is it true in general? Does anybody have a counter example? Or a proof?


When measure is infinite then the following way can be used:

Let $$\mathcal S:=\left\{A\in \mathcal{B}\mid \forall\varepsilon>0,\exists A'= \bigcup_{n=1}^\infty A_n \text{ such that } A_n \in \mathcal{A} \text{ and }\mu(A\Delta A')\leq \varepsilon\right\}.$$

Note that $\mathcal A \subset \mathcal S$ and $\mathcal S$ satisfies the properties of $\sigma$ algebra. Thus $\mathcal S = \sigma(\mathcal{A})=\mathcal{B}$.

Now $A\in \mathcal{B}$ is such that $\mu(A)< \infty$ then truncate $A'$ appropriately to get the result.


If we consider

$$\mathcal S':=\left\{A\in \mathcal{B}\mid \forall\varepsilon>0,\exists A'= \bigcup_{n=1}^\infty A_n, B'= \bigcup_{n=1}^\infty B_n \text{ such that } A_n, B_n\in \mathcal{A} \text{ and }\mu(A\Delta A')\leq \varepsilon, \mu(A^c\Delta B')\leq \varepsilon\right\}.$$

then it would be easier to check that $\mathcal S'$ is a closed under complementation. Check that $\mathcal S'$ is a $\sigma$ algebra. It is easy to see that $\mathcal A \subset \mathcal S'$ and $\mathcal S' = \sigma(\mathcal{A})=\mathcal{B}$. Now again $A\in \mathcal{B}$ is such that $\mu(A)< \infty$ then truncate $A'$ appropriately to get the result.

However, note that, since $\mathcal S'\subset \mathcal S$, $\mathcal S$ has to be closed under complementation.

  • $\begingroup$ How do you check that $\mathcal{S}$ is stable by complementation? It seems that in general the complement of a countable union of elements of $\mathcal{A}$ is not a countable union of elements of $\mathcal{A}$: take $[0,1]$ with $\mathcal{A}$ being all finite disjoint unions of open intervals and take $A'$ to be the complement of the triadic Cantor set. $\endgroup$
    – Gagar
    Dec 6, 2020 at 11:30
  • $\begingroup$ @Gagar Thanks for your comment. I have edited the answer. $\endgroup$
    – Mayuresh L
    Dec 6, 2020 at 13:49
  • $\begingroup$ But now how do you check that $\mathcal{S}'$ is closed under countable unions? $\endgroup$
    – Gagar
    Dec 6, 2020 at 18:09
  • $\begingroup$ @Gagar Just as in the answer by Davide Giraudo: first show it for finite union and then for countable pairwise disjoint union. $\endgroup$
    – Mayuresh L
    Dec 6, 2020 at 19:30
  • $\begingroup$ I agree for the part $\mu(A \Delta A')$ but what about the part $\mu(A^c \Delta B')$? Could you please give some more detail? $\endgroup$
    – Gagar
    Dec 6, 2020 at 19:41

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