Let $\{E_k\}^{\infty}_{k=1}$ be a countable disjoint collection of measurable sets. Prove that for any set A, $m^*(A \cap \cup^{\infty}_{k=1})$ = $\sum^{\infty}_{k=1} m^*(A\cap E_k)$

My answer:

I tried to prove this by induction:

We know that if k=1, $m^*(A \cap E) = m^*(A \cap E)$. So now if we assume that for some n, $m^*(A \cap \cup^{n}_{k=1})$ = $\sum^{n}_{k=1} m^*(A\cap E_k)$. Now we have to prove that it is true for n+1. $m^*(A \cap \cup^{n+1}_{k=1}) \leq m^*((A \cap \cup^{n}_{k=1} E_k) + m^*(A \cap E_{n+1})$

So now I have to prove the reverse inclusion, right?...But I'm kind of stuck here...

  • 1
    $\begingroup$ And you should also use that $E_k$ are measurable. $\endgroup$ – Berci Feb 11 '13 at 15:27

I assume $m^*$ is an outer measure; and will use the fact that a set $E$ is measurable if and only if
$$m^*(E)=m^*(E\cap T)+m^*(E\cap T')$$ for all sets $T$.

Since $\cup_{k=1}^\infty E_k$ is measurable, your result is equivalent to proving that $$\tag{1} m^*(A)=\sum_{k=1}^\infty m^*(E_k\cap A) +m^*\Bigr(A\cap \bigl(\ \bigcup_{k=1}^\infty E_k\ \Bigr)'\Bigr). $$

Towards that end, note, by the the countable subadditivity of $m^*$ $$\eqalign{ m^*(A) &\le m^*\Bigr(A\cap \Bigl(\ \bigcup_{k=1}^\infty E_k\ \Bigr) \Bigr) +m^*\Bigr(A\cap \Bigl(\ \bigcup_{k=1}^\infty E_k\ \Bigr)'\Bigr)\cr &\le\sum_{k=1}^\infty m^*(E_k\cap A) +m^*\Bigr(A\cap \Bigl(\ \bigcup_{k=1}^\infty E_k\ \Bigr)'\Bigr). } $$

With this in hand, we see that the desired result is trivial if $m^*(A)=\infty$.

Now assume $m^*(A)\ne\infty$. To prove the reverse inequality needed, we will prove by induction that for each $p\ge1$ and any set $A$ $$\tag{2} m^*(A)=\sum_{k=1}^p m^*(E_k\cap A) +m^*\Bigr(A\cap \Bigl(\ \bigcup_{k=1}^p E_k \Bigr)'\ \Bigr). $$ Once this is done, since the sequence $\Bigl(m^*\Bigr(A\cap \bigl(\cup_{k=1}^p E_k \bigr)'\Bigr)\Bigr)_{p=1}^\infty$ is nonincreasing and bounded below by $m^*\Bigr(A\cap \bigl(\cup_{k=1}^\infty E_k \bigr)' \Bigr)$, it will follow that $$ m^*(A)\ge\sum_{k=1}^\infty m^*(E_k\cap A) +m^*\Bigr(A\cap \Bigl(\ \bigcup_{k=1}^\infty E_k\ \Bigr)'\Bigr), $$ as desired.

So, on to the proof of $(2)$.

The base case follows from the fact that $E_1$ is measurable: $$ m^*(A)=m^*(A\cap E_1)+m^*(E\cap E_1'). $$

Assume $(2)$ is true for $p=k$.

Since $E_{p+1}$ is measurable, $$m^*(A)=m^*(A\cap E_{p+1})+ m^*(A\cap E_{p+1}')$$ Now, using $(2)$ applied to the set $A\cap E_{p+1}'$,

$$ \eqalign{ m^*(A)&=m^*(A\cap E_{p+1})+ m^*(A\cap E_{p+1}')\cr &=m^*(A\cap E_{p+1})+\sum_{k=1}^p m^*(A\cap E_{p+1}' \cap E_k) +m^*\Bigl( A\cap E_{p+1}'\cap \Bigl( \bigcup_{k=1}^p E_k\ \Bigr)'\Bigr)\cr &=m^*(A\cap E_{p+1})+\sum_{k=1}^p m^*(A\cap E_k) +m^*\Bigl( A\cap E_{p+1}'\cap \Bigl(\ \bigcup_{k=1}^p E_k\ \Bigr)'\Bigr)\cr &=\sum_{k=1}^{p+1} m^*(E_k\cap A) +m^*\Bigr(A\cap \Bigl(\ \bigcup_{k=1}^{p+1} E_k\ \Bigr)'\Bigr); } $$ where, in the third equality, we used the fact that ${E_k\subset E_{p+1}'}$. This is exactly what we need to finish the proof by induction. $(2)$ holds for all $p$.


First, there's philosophical problems with using an induction argument to prove the infinite property you're trying to show. So I don't think that will work. Additionally, I found David Mitra's solution to the problem to be rather complicated compared to the one that follows.

Here goes nothing.

By countable subadditivity of outer measure, \begin{align} m^*\left(A\cap\bigcup_{k=1}^\infty E_k\right)&=m^*\left(\bigcup_{k=1}^\infty (A\cap E_k)\right)\\ &\leq \sum_{k=1}^\infty m^*(A\cap E_k), \end{align} so it really suffices to prove the opposite inequality, i.e., $\sum_{k=1}^\infty(A\cap E_k)\leq m^*\left(A\cap\bigcup_{k=1}^\infty E_k\right)$.

Since $A\cap\bigcup_{k=1}^\infty E_k$ contains $A\cap\bigcup_{k=1}^n E_k$ for each $n$, we have: \begin{align} m^*\left(A\cap\bigcup_{k=1}^\infty E_k\right)&\geq m^*\left(A\cap\bigcup_{k=1}^n E_k\right)\\ &=\sum_{k=1}^n m^*(A\cap E_k)\tag{1} \end{align} by Proposition 6 of Royden $\S 2.3$. This gives: \begin{align} \lim_{n\to\infty} m^*\left(A\cap\bigcup_{k=1}^\infty E_k\right)&\geq \lim_{n\to\infty} \sum_{k=1}^n m^*(A\cap E_k)\\ m^*\left(A\cap\bigcup_{k=1}^\infty E_k\right)&\geq\sum_{k=1}^\infty m^*(A\cap E_k), \end{align} since the LHS is independent of $n$. Q.E.D.


I don't think you can use induction to prove a property for $\infty$.

Why not using that measures are completely additive and the equality $A\cap \bigcup_{k=1}^\infty E_k = \bigcup_{k=1}^\infty(A\cap E_k)$

  • $\begingroup$ Actually, this is what I thought of doing before I used induction... $m^*(A\cap \bigcup_{k=1}^\infty E_k) = m^*(\bigcup_{k=1}^\infty(A\cap E_k))$ and we know that, since each of these is countable and disjoint we can say that it is equal to $\sum^{\infty}_{k=1}m(A \cap E_k)$ But before I use this theorem, I need to show that $A \cap E_k$ is measurable for each k, right? But there is a theorem in my textbook that says that we cannot conclude that a subset of a measurable set is measurable. $\endgroup$ – user58289 Feb 11 '13 at 16:12
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    $\begingroup$ Since $A\cap E_k$ is a subset of $\{E_k\}^{\infty}_{k=1}$, we cannot say taht $A \cap E_k$ is measurable either, right? That's what confused me... $\endgroup$ – user58289 Feb 11 '13 at 16:13

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