What is the difference between outer measure and Lebesgue measure?

We know that there are sets which are not Lebesgue measurable, whereas we know that outer measure is defined for any subset of $\mathbb{R}$.

  • $\begingroup$ Hi, lavy good to see you here.Guess MATH.Se is working there now. $\endgroup$
    – anonymous
    Mar 13, 2011 at 9:04

3 Answers 3


The Lebesgue measure and Lebesgue outer measure coincide on Lebesgue measurable sets, which can be defined in several equivalent ways. Let $m$ and $m^*$ denote the Lebesgue measure and the Lebesgue outer measure respectively. These are some possible definitions of $A\subset\mathbb{R}^n$ being measurable:

  1. For all $B\subset\mathbb{R}^n$ $$ m^*(B)=m^*(B\cap A)+m^*(B\setminus A) $$
  2. For all $\epsilon>0$ there exist an open set $G$ and a closed set $F$ such that $F\subset A\subset G$ and $m^*(G\setminus F)<\epsilon$. (Note that since $G\setminus F$ is open, it is measurable, so that $m^*(G\setminus F)=m(G\setminus F)$.)
  3. $A=F\cup N$, where $F$ is an $F_\sigma$ (i.e. a countable union of closed sets) and $m(N)=0$.
  4. $A=G\setminus N$, where $G$ is a $G_\delta$ (i.e. a countable intersection of open sets) and $m(N)=0$.

The reason for the need of two different concepts is that neither of them is "perfect":

  • $m$ is a measure, but is not defined for all subsets of $\mathbb{R}^n$
  • $m^*$ is defined for all subsets of $\mathbb{R}^n$, but is not additive: here exist disjoint sets $A$ and $B$ such that $m^*(A\cup B)\ne m^*(A)+m^*(B)$.
  • 3
    $\begingroup$ I am also studying the difference between outer measure and Lebesgue measure. I can't seem to think of a good example in finding two disjoint sets $A$ and $B$ that satisfy the subadditivity property; do you happen to know of a simple example of such two disjoint sets? $\endgroup$
    – Cookie
    Mar 2, 2015 at 6:55
  • $\begingroup$ Have a look at this question. $\endgroup$ Mar 2, 2015 at 10:37
  • $\begingroup$ "Here exists disjoint sets $A$ and $B$ such that m*($A∪B$) ≠ m*($A$) + m*($B$)"--can you help me find such an example? Thanks for your help. $\endgroup$ Apr 15, 2015 at 15:25

As a supplement to Julián Aguirre's answer, note that Lebesgue originally wanted a measure $m$ to satisfy certain properties:

  1. $m(S)>0$ for any set $S$
  2. $m$ is identical to length when considering intervals
  3. $m$ is translation invariant, i.e. if you slide your set up or down the real line, its measure should be unchanged
  4. $m$ should be (countably) additive.

Now, Lebesgue originally introduced both inner and outer measure, which were (respectively) under and over estimates of a set's true "size", but these fail to be countably additive. Instead of trying to find some new measure which satisfies all 4 properties, he restricted to a smaller collection of sets (as in Julián's answer) called "measurable sets" for which outer measure does satisfy 1-4.

This was a smart move, since it turns out that there is no nontrivial function satisfying 1-4 for every subset of $\mathbf{R}$.


Lebesgue outer measure (m*) is for all set E of real numbers where as Lebesgue measure (m) is only for the set the set of measurable set of real numbers even if both of them are set fuctions.

by Geleta Tadele Mohammed


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