If $E \subset \mathbb{R}$ and $r > 0$, we define $rE = \{rx : x \in E \}$. If $\mu^*(E) = l$, find $\mu^*(rE)$

Question

I'm learning about Measure Theory (specifically Lebesgue outer measure) and need some help to understand the solution to this problem:

If $E \subset \mathbb{R}$ and $r > 0$, we define $rE = \{rx : x \in E \}$.

If $\mu^*(E) = l$ (where $\mu^*(E)$ is the Lebesgue outer measure of $E$), find $\mu^*(rE)$.

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I should mention that this problem has already been asked by other members but I could not find any answer to my questions. In my effort to solve this problem I came across this statement:

(1) : If $E$ is any set, measurable or not, then $\mu^*(rE) = r\mu^*(E)$ for all $r > 0$.

Taking into consideration the proposition (1) then the solution is immediate.

Since $\mu^*(E) = l$ and $\mu^*(rE) = r\mu^*(E)$ for all $r > 0$ then

$$\mu^*(rE) = rl \;\; \text{for all} \ r > 0.$$

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Intuitively this solution makes sense but I don't understand why the equality (1) holds. Is there a simple way to prove it? Furthermore, I don't understand why we can't have the below inequality instead:

$$\mu^*(rE) \leq r\mu^*(E) \;\; \text{for all} \ r > 0.$$

The problem doesn't say anything about the set $E$ apart from the fact that it is a subset of $\mathbb{R}$. The set $E$ could be the union of countable sets which overlap each other so it would make sense to have an inequality instead.

Answer 1

In general, whenever we deal with measures (or integrals, or things of the sort), there is some kind of simple sets (or functions) which generate all other things, in some sense. You can see this, for example, when we define integrals: First define integrals for characteristic functions, then for simple functions, and then take some limits and define for arbitrary positive functions.

So a general rule is that we should always try to simplify our problems to a simpler case, by taking some generating elements.

In your specific problem: We are dealing with outer Lebesgue measure. The outer Lebesgue measure is defined in the following way: For any $E\subseteq\mathbb{R}$: $$\mu^*(E)=\inf\left\{\sum_{i=1}^\infty\mu(I_i):I_i\text{ intervals, }E\subseteq\bigcup_{i=1}^\infty I_i\right\},$$ where $\mu((a,b))=b-a$ (where $a\leq b$) is the usual interval length. (This is simply the definition. Maybe in class/book you consider only closed/open intervals, or something of the sort, but the arguments work the same in either case.)

So in this case, intervals work as some sort of "generating elements", and this means, more or less, that it is enough to prove that $\mu^*(rI)=r\mu^*(I)$ for all intervals $I$. But this is obvious.

So, to prove that $\mu^*(rE)=r\mu^*(E)$, let's use the definitions: Suppose $I_i$ are intervals with $E\subseteq\bigcup I_i$. Then $rI_i$ are intervals and $rE\subseteq r\bigcup I_i=\bigcup rI_i$, so the definition of $\mu^*(rE)$ yields $$\mu^*(rE)\leq\sum_{i=1}^\infty\mu(rI_i)=r\sum_{i=1}^\infty\mu(I_i)$$ Taking the infimum on the right, for all collections of intervals which cover $E$, we obtain $\mu^*(rE)\leq r\mu^*(E)$.

Now you should try to prove the converse inequality, using the same kind of arguments, but with the roles of $E$ and $rE$ switched, and $1/r$ in place of $r$. I will leave an alternative, faster way to do it below:

! From the previous argument, we have $$\mu^*(rE)\leq r\mu^*(E)\qquad\text{ for all }r>0\text{ and }E\subseteq\mathbb{R}$$ Since this is valid for all $r$ and $E$, we apply this with $1/r$ in place of $r$ and $rE$ in place of $E$, obtaining $$\mu^*(E)=\mu^*(\frac{1}{r}(rE))\leq\frac{1}{r}\mu^*(rE)$$ so $\mu^*(rE)\geq r\mu^*(E)$.