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Prove directly from the definition of Outer Measure $m^{*}$ that if $m^{*}(A) = 0$, then $m^{*}(A \cup B)=m^{*}(B)$ (this is actually an exercise from Royden, but our professor wants us to use the definition of Outer Measure (a function $m^{*}: \{\text{subsets of}\,\mathbb{R}\} \to \mathbb{R}_{\geq 0} \cup \{\infty \}$ defined as:

$m^{*}(A) = \inf \left \{\displaystyle \sum_{k=1}^{\infty} l(I_{k})\, \vert I_{1},I_{2},\cdots \text{open bounded intervals such that}\, A \subseteq \cup_{k=1}^{\infty} I_{k} \right\}$ (where $l(I_{k})$ is the length of the $k$th interval)).

Now, let $m^{*}(B) = \inf \left \{\displaystyle \sum_{k=1}^{\infty} l(J_{k})\, \vert J_{1},J_{2},\cdots \text{open bounded intervals such that}\, B \subseteq \cup_{k=1}^{\infty} J_{k} \right\}$ and $m^{*}(A \cup B) = \inf \left \{\displaystyle \sum_{k=1}^{\infty} l(M_{k})\, \vert M_{1},M_{2},\cdots \text{open bounded intervals such that}\, A \cup B \subseteq \cup_{k=1}^{\infty} M_{k} \right\}$

The way I think the proof should go is the following:

$m^{*}(A \cup B) = \inf \sum_{k=1}^{\infty} l(M_{k}) = \inf \sum_{k=1}^{\infty}l(I_{k}) + \inf \sum_{k=1}^{\infty}l(J_{k}) = 0 + m^{*}(B) = m^{*}(B)$.

However, in order to do this, I would (1) need to show that $\cup_{k=1}^{\infty} M_{k} = (\cup_{k=1}^{\infty}I_{k})\cup (\cup_{k=1}^{\infty}J_{k})$ and (2) that this implies that $\inf \sum_{k=1}^{\infty} l(M_{k}) = \inf\sum_{k=1}^{\infty}l(I_{k}) + \inf\sum_{k=1}^{\infty}l(J_{k})$.

It's frustrating, too, because this is supposed to be an easy exercise, and intuitively, it makes sense that the absolute minimum needed to cover $A \cup B$ should also be the absolute minimum needed to cover $A$ plus the absolute minimum needed to cover $B$, but I need to show this in a mathematically rigorous way, and can't just assert things without being able to mathematically explain them.

If someone could please help me fill in the justification gaps in this proof and/or fix it if it's incorrect, it would be very much appreciated, and would definitely help me in my attempts to tackle the harder problems.

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  • $\begingroup$ You have to prove $m*(A\cup B)\ge m*(B)$ and $m*(A\cup B)\le m*(B).$ Which inequality are you stuck on? $\endgroup$
    – bof
    Oct 10, 2015 at 23:40
  • $\begingroup$ $m^{*}(A \cup B) \leq m^{*}(B)$. That's really all I need to show here? What about showing that $\inf \sum_{k=1}^{\infty} l(M_{k}) = \inf \sum_{k=1}^{\infty}l(I_{k}) + \inf \sum_{k=1}^{\infty}l(J_{k})$? How do I prove that? $\endgroup$
    – user100463
    Oct 10, 2015 at 23:40
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    $\begingroup$ Do you know how to prove from the definition of outer measure that $m^*(A\cup B)\le m^*(A)+m^*(B)$? Doesn't the proof of that general fact work in the special case where $m^*(A)=0$? $\endgroup$
    – bof
    Oct 10, 2015 at 23:43
  • $\begingroup$ To show $m^*(A\cup B)\le m^*(B)$ it's enough to show that, for every $\varepsilon\gt0$, you have $m^*(A\cup B)\le m^*(B)+\varepsilon.$ $\endgroup$
    – bof
    Oct 10, 2015 at 23:46
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    $\begingroup$ I'm going to try to prove, using the definition of outer measure, that $m^{*}(A \cup B) \leq m^{*}(A)+m^{*}(B)$. I'll let you know how it goes! Thanks :) $\endgroup$
    – user100463
    Oct 10, 2015 at 23:48

2 Answers 2

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My solution doesn't use the definition of the outer measure, but it goes as follows:

First, since $m^*(A) =0$. Then, $$m^*(A \cup B) \leq m^*(A) + m^*(B) \leq m^*(B).$$ It suffices to show that $m^*(B) \leq m^*(A \cup B)$. We need to consider three cases.


If $A = \emptyset$, then $A \cup B = B$ and $m^*(B) \leq m^*(A \cup B) = m^*(B)$.


If $\emptyset \neq A \subseteq B$, then $A \cup B = B$ and $m^*(B) \leq m^*(A \cup B) = m^*(B)$.


If $\emptyset \neq A \nsubseteq B$, then $B \subset A \cup B$ and $m^*(B) \leq m^*(A \cup B)$.


As you can see all the three cases gives $m^*(B) \leq m^*(A \cup B)$ as required.

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If $B$ has infinite outer measure, the equality holds trivially.

So, suppose instead that $m^{*}(B) < \infty$.

For $B$, let $\epsilon > 0$. Then, $\exists$ a countable collection $\{I_{B,i}\}_{i=1}^{\infty}$ of open, bounded intervals for which $B \subseteq \sum_{i=1}^{\infty}I_{B,i}$, and $\sum_{i=1}^{\infty}l(I_{B,i})<m^{*}(B) + \epsilon/2$ (since, by definition of $m^{*}$ as an outer measure, $m^{*}(B) = \inf \sum_{i=1}^{\infty}l(I_{B,i})$, $m^{*}(B)+\epsilon/2$ cannot be a lower bound for $\sum_{i=1}^{\infty}l(I_{B,i})$, so we must have that $\sum_{i=1}^{\infty}l(I_{B,i})<m^{*}(B) + \epsilon/2$.

Similarly, for $A$, we have that for $\epsilon > 0$, $\sum_{i=1}^{\infty}l(I_{A,i}) < m^{*}(A)+\epsilon/2 < 0 + \epsilon/2 = \epsilon/2$.

Now, for greater ease of notation, let $A:=E_{1}$ and $B:=E_{2}$. Then, we have $m^{*}\left( \sum_{k=1}^{2}E_{k}\right) = \inf \left\{\sum_{1\leq k \leq 2, \,i < \infty}l(I_{k,i}) \right\} = \sum_{1\leq k \leq 2, \,i < \infty}l(I_{k,i}) = \sum_{i=1}^{\infty} l(I_{1},i) + \sum_{i=1}^{\infty} l(I_{2},i) < (m^{*}(E_{1})+\epsilon/2) + (m^{*}(E_{2})+\epsilon/2) = (0 + \epsilon/2) + (m^{*}(E_{2})+\epsilon/2) = m^{*}(E_{2}) + \epsilon$.

Since this holds $\forall \epsilon > 0$, it also holds for $\epsilon = 0$.

Now, we must show that $m^{*}(B) = 0 + m^{*}(B) = m^{*}(A) + m^{*}(B) \leq m^{*}(A \cup B)$.

For ease of notation, again let $A = E_{1}$, $B=E_{2}$. Then, $m^{*}(E_{2})=0+m^{*}(E_{2}) = m^{*}(E_{1}) + m^{*}(E_{2}) = \inf \left\{\sum_{i=1}^{\infty}l(I_{1,i}) \right\} + \inf \left\{\sum_{i=1}^{\infty}l(I_{2,i}) \right\} \leq \sum_{i=1}^{\infty} l(I_{1,i}) + \sum_{i=1}^{\infty}l(I_{2,i}) = \sum_{i=1}^{\infty}\left[ l(I_{1,i}) + l(I_{2,i})\right] = \sum_{1 \leq j \leq 2,\,i<\infty}l(I_{k,i}) $

Now, $m^{*}(E_{1} \cup E_{2}) = \inf \left\{\sum_{1 \leq j \leq 2,\,i<\infty}l(I_{k,i}) \right\}$, so $m^{*}(E_{1} \cup E_{2})+\epsilon$, where $\epsilon > 0$, cannot be a lower bound for $\sum_{1 \leq j \leq 2,\,i<\infty}l(I_{k,i})$.

Thus, our inequality above becomes $< m^{*}(E_{1}\cup E_{2})+\epsilon$.

Since this holds $\forall \epsilon > 0$, it also holds for $\epsilon = 0$.

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  • $\begingroup$ @bof, what do you think? $\endgroup$
    – user100463
    Oct 11, 2015 at 14:54

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