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I have proved the following statement and I would like to know If I have made any mistakes, thanks.

"Suppose $A\subset\mathbb{R}$ and $t\in\mathbb{R}$. Let $tA:=\{ta:a\in A\}$. Prove that $|tA|=|t||A|$. (Assume that $0\cdot\infty$ is defined to be $0$)."


NOTE: $|\cdot|$ refers to outer measure, i.e. for $A\subset\mathbb{R},\ |A|:=\inf\{\sum_{k=1}^{\infty}l(I_k): I_1,I_2,\dots\text{ are open intervals such that }A\subset\bigcup_{k=1}^{\infty}I_k\}$; the length of an open interval $I\subset\mathbb{R}$ is defined as

$\ell(I):=\begin{cases} b-a & \text{if }I=(a,b),\ a,b\in\mathbb{R}, a<b; \\ 0 & \text{if }I=\emptyset; \\ \infty & \text{if } I=(-\infty, a)\text{ or } I=(a,\infty);\\ \infty & \text{ if }I=(-\infty,\infty) \end{cases} $


My proof: If $t=0$ then $tA=\{0\}$ so $|tA|=0$ (since finite sets have outer measure $0$) and $|t||A|=\begin{cases} 0\cdot |A| =0 & \text{ if } |A|<\infty\\ 0\cdot |A| =0 & \text{ if } |A|=\infty & \text{(since by hypothesis }0\cdot\infty=0) \end{cases}\ =0$ so $|tA|=|t||A|$, as desired.

Now, suppose $t\neq 0$ and let $I_1,I_2,\dots$ be a sequence of open intervals ($I_k=(a_k,b_k),\ a_k\in\mathbb{R}\cup-\{\infty\}, b_k\in\mathbb{R}\cup \{+\infty\}, a_k<b_k$ or $I_k=\emptyset$) such that $A\subset\bigcup_{k=1}^{\infty}I_k$: then $tI_1,tI_2,\dots $is a sequence of open intervals such that $tA\subset\bigcup_{k=1}^{\infty}tI_K$ ($a\in A\Rightarrow a\in I_K=(a_K,b_K)\ a_K<b_K$ for some $K\geq 1\Rightarrow a_K<a<b_K\Rightarrow \begin{cases} ta_K<ta<tb_K & \text{ if } t>0 \\ tb_K<ta<ta_K &\text{ if } t<0 \end{cases}$ i.e. $ta\in tI_K$) and $$l(tI_k):=\begin{cases} tb_k-ta_k & \text{if }I_k=(a_k,b_k),\ a_k,b_k\in\mathbb{R}, a_k<b_k, t>0; \\ ta_k-tb_k & \text{if }I_k=(a_k,b_k),\ a_k,b_k\in\mathbb{R}, a_k<b_k, t<0; \\ 0 & \text{if }I_k=\emptyset; \\ \infty & \text{if } I_k=(-\infty, b_k)\text{ or } I_k=(a_k,\infty);\\ \infty & \text{ if }I_k=(-\infty,\infty) \end{cases}= \begin{cases} t(b_k-a_k) & \text{if }I_k=(a_k,b_k),\ a_k,b_k\in\mathbb{R}, a_k<b_k, t>0; \\ -t(b_k-a_k) & \text{if }I_k=(a_k,b_k),\ a_k,b_k\in\mathbb{R}, a_k<b_k, t<0; \\ 0 & \text{if }I_k=\emptyset; \\ \infty & \text{if } I_k=(-\infty, b_k)\text{ or } I_k=(a_k,\infty);\\ \infty & \text{ if }I_k=(-\infty,\infty) \end{cases}=\begin{cases} |t|(b_k-a_k) & \text{if }I_k=(a_k,b_k),\ a_k,b_k\in\mathbb{R}, a_k<b_k, t\neq 0; \\ |t| 0 & \text{if }I_k=\emptyset; \\ |t|\infty & \text{if } I_k=(-\infty, b_k)\text{ or } I_k=(a_k,\infty);\\ |t|\infty & \text{ if }I_k=(-\infty,\infty) \end{cases}=|t|\begin{cases} (b_k-a_k) & \text{if }I_k=(a_k,b_k),\ a_k,b_k\in\mathbb{R}, a_k<b_k; \\ 0 & \text{if }I_k=\emptyset; \\ \infty & \text{if } I_k=(-\infty, b_k)\text{ or } I_k=(a_k,\infty);\\ \infty & \text{ if }I_k=(-\infty,\infty) \end{cases}=|t|I_k$$ so $$|tA|\leq\sum_{k=1}^{\infty}\ell (tI_k)=\sum_{k=1}^{\infty}|t|\ell (I_k)=|t|\sum_{k=1}^{\infty}\ell (I_k)$$ and by taking the $\inf$ over all open coverings of $A$ we find that $$|tA|\leq |t||A|$$ From this inequality we also get $$\left| \frac{1}{t}tA \right| \leq \left| \frac{1}{t} \right||tA|\Rightarrow |A|\leq \frac{1}{|t|}|tA|\Rightarrow |t||A|\leq |tA|$$ thus $|tA|=|t||A|\ \forall t\in\mathbb{R}, A\subset\mathbb{R}$, as desired.

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2 Answers 2

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You have to be a bit careful declaring the equality of the infima at the end of the proof. They are of course equal, but that is essentially what you need to prove.

You've established that if $I$ is an interval and $t \in \mathbf R$ then $\ell(tI) = |t| \ell(I)$. If $A \subset \mathbf R$ and $\{I_k\}$ is a countable cover of $A$ by intervals, then $\{tI_k\}$ is a countable cover of $tA$ and thus $$|tA| \le \sum \ell(tI_k) = |t| \sum \ell(I_k).$$ Now take the infimum over all countable coverings of $A$ to find $$|tA| \le |t||A|.$$

This inequality also implies $$ \left| \frac 1t tA \right| \le \frac{1}{|t|} |tA|$$ whenever $t \not= 0$ giving you $|t||A| \le |tA|$ as well.

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  • $\begingroup$ Brilliant! I have edited my question according to your correction. Thank you very much. $\endgroup$
    – lorenzo
    Mar 30, 2021 at 9:55
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I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
This exercise is Exercise 2 on p.23 in Exercises 2A in this book.

You corrected your original question, so I think there is no mistake in your question.

I have imitated Axler's proof of the following proposition.  

2.7 outer measure is translation invariant on p.16
Suppose $t\in\mathbb{R}$ and $A\subset\mathbb{R}$. Then $|t+A|=|A|$.

My proof:
If $t=0$, then $tA=\{0\}$.
By 2.3 (or 2.4) on p.15, $|tA|=0$.
If $|A|=\infty$, then $|t||A|=0$ by the assumption in Exercise 2.
If $|A|<\infty$, then of course, $|t||A|=0$.
So, if $t=0$, then $|tA|=|t||A|$ holds.

We conseider the case in which $t\neq 0$.
Suppose $I_1,I_2,\dots$ is a sequence of open intervals whose union contains $A$. Then $tI_1,tI_2,\dots$ is a sequence of open intervals whose union contains $tA$.
Thus $$|tA|\leq\sum_{k=1}^{\infty}l(tI_k)=\sum_{k=1}^{\infty}|t|l(I_k)=|t|\sum_{k=1}^{\infty}l(I_k),$$
$$\frac{1}{|t|}|tA|\leq\sum_{k=1}^{\infty}l(I_k).$$ Taking the infimum of the last term over all sequences $I_1,I_2,\dots$ of open intervals whose union contains $A$, we have $\frac{1}{|t|}|tA|\leq |A|$. So, $|tA|\leq |t||A|$.

To get the inequality in the other direction, note that $A=\frac{1}{t}(tA)$. Thus applying the inequality from the previous paragraph, with $A$ replaced by $tA$ and $t$ replaced by $\frac{1}{t}$, we have $|A|=|\frac{1}{t}(tA)|\leq\left| \frac{1}{t} \right||tA|$. So, $|t||A|\leq |tA|$. Hence $|tA|=|t||A|$.

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