Outer measure (Lebesgue outer measure)


Lebesgue outer measure satisfies the inequality $m^* (A \cup B) + m^* (A \cap B) \le m^* (A) +m^* (B)$. Substitute to get the desired result.

The above inequality should not be difficult to prove no matter how you define Lebesgue outer measure. Basically it's an approximation argument. For instance, if you already know that $m^* (A)$ is the infimum of $m(G)$ over all open sets $G$ containing $A$, then you just pick an $\varepsilon >0$, take open sets $G_A$ and $G_B$ which $\varepsilon$-approximate $A$ and $B$ (in the above sense) and then $G_A \cup G_B$ and $G_A \cap G_B$ will approximate well $A \cup B$ and $A \cap B$ (respectively). Since Lebesgue measure is additive on open sets (and more generally, on measurable sets), the inequality will then follow in a straightforward manner.

Similarly, you can prove that Lebesgue inner measure satisfies the reverse inequality.


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Yeonjoo Yoo
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Yeonjoo Yoo

Updated on August 14, 2020


  • Yeonjoo Yoo
    Yeonjoo Yoo over 3 years

    Let $A,B \subset \mathbb{R}$ and $m^*(A)=m^*(B)=1$ and $m^*(A\cup B)=2$. Prove that $m^*(A\cap B)=0$.

    I tried every way I can think of but I do not know how to figure this out.

    Only properties that I am aware of are monotonicity, countable subadditivity, the outer measure of empty set is zero, translation invariant.

    Thank you in advance.

    • Jonas Meyer
      Jonas Meyer over 12 years
      Have you been introduced to measurable sets?
    • Mark
      Mark over 12 years
      The outer measure satisfies the inequality $m^* (A \cup B) + m^* (A \cap B) \le m^* (A) +m^* (B)$. Substitute.
    • user1736
      user1736 over 12 years
      @Mark: I know that the OP's question is referring to Lebesgue measure, but is that inequality also only true for Lebesgue measure?
    • LostInMath
      LostInMath over 12 years
      @user1736: I think the inequality does not hold for general outer measures. For example consider $X=\{0,1,2\}$ and the function $\mu:\mathcal{P}(X)\to[0,\infty]$ with $\mu(\emptyset)=0$, $\mu(A)=1$, if $A\in\mathcal{P}(X)\setminus\{\emptyset,X\}$ and $\mu(X)=2$. If I'm not mistaken, $\mu$ is an outer measure on $X$ but, choosing $A=\{0,1\}$ and $B=\{1,2\}$, we see that the inequality does not hold.
    • t.b.
      t.b. over 12 years
      @LostInMath: I haven't checked your example but you're right, it doesn't hold for general outer measures. It holds for regular outer measures and the property is called strong subadditivity or submodularity. For Lebesgue measure it holds and is not particularly hard to show.
    • user1736
      user1736 over 12 years
      @Theo and LostinMath: Thanks! I verified the result using approximation with Borel sets, and I wasn't sure how else to do it. I guess being regular is the actual property that I was using.
    • t.b.
      t.b. over 12 years
      @user1736 that's definitely the way to go.