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Background information: $L$ is the class of Lebesgue measurable sets. $m$ is the Lebesgue measure which is a complete measure $\mu_F$ associated to the function $F(x) = x$, for which the measure of an interval is simply its length. $\mu_F$ is the Lebesgue-Stieltjes measure associated with $F$.

Theorem 1.21 - If $E\in L$ then $E+s\in L$ and $rE\in L$ for all $s,r\in\mathbb{R}$. Moreover, $m(E + s) = m(E)$ and $m(rE) = |r|m(E)$.

Proof - Let $E\subset\mathbb{R}$. If $E\in L$ and $r,s\in\mathbb{R}$ define $$E + s = \{x + s: x\in E\}$$ and define $$rE = \{rx:x\in E\}$$ We need to show that if $E\in L$, then $E + s\in L$ and $rE\in L$. For all $Y\subset \mathbb{R}$ $$m^*(E) = m^*(E\cap Y) + m^*(E\cap Y^c)$$ Let $E = E - s$ then $$m^*(E) = m^*(E-s) = m^*((E-s)\cap Y) + m^*((E-s)\cap Y^c)$$

I am not really sure if this is on the right track and I do not understand how to show the Moreover part. Any suggestions is greatly appreciated.

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It's maybe too late to give an answer now, but since I was stuck on the same theorem a few days ago, I think I should give a detailed answer, using Folland's proof.

Recall that the class $\mathcal{L}$ of Lebesgue measurable sets is just the completion of the Borel $\sigma$-algebra $\mathcal{B}_\mathbb{R}$, which means that $$\mathcal{L}=\big\{E\cup F:E\in \mathcal{B}_\mathbb{R} \text{ and } F\subseteq N \text{ for some } N\in \mathcal{B}_\mathbb{R} \text{ such that } m(N)=0\big\}.$$

Let $A \in \mathcal{L}$ and $s, \lambda \in \mathbb{R}$. We want to prove that $A+s\in \mathcal{L}$ and $\lambda A \in \mathcal{L} $. Using the definiton of $\mathcal{L}$, we can find a Borel set E and a set $F\subseteq N$ for some $N\in \mathcal{B}_\mathbb{R}$ with $m(N)=0$ such us $A=E\cup F$. We can assume that E and F are disjoint (otherwise we can replace E, F and N with E, $F\setminus E$ and $N\setminus E$). In oder to prove that $A+s\in \mathcal{L}$ and $\lambda A \in \mathcal{L} $, we just have to show that $A+s$ and $\lambda A$ are unions of a Borel set and a Lebesgue null set. Since \begin{equation}\tag{1} A+s=(E\cup F)+s=(E+s)\cup(F+s) \text{ and } \lambda A=\lambda E \cup \lambda F \end{equation} we need to prove that $E+s,\lambda E$ are Borel sets and $F+s,\lambda F$ are Lebesgue null sets.

It is a fact that Borel sets are invariant under translations and dilations. For a proof, use this link: translation and dilation invariance of borel sets. It follows that $E+s,\lambda E$ are Borel sets.

Now define the measures $m_s(B):=m(B+s)$, $m^\lambda(B):=m(\lambda B)$ for every $B\in \mathcal{L}$ such that $B+s,\lambda B \in \mathcal{L}$. Observe that $m_s$ and $m^\lambda$ agree with $m$ and $|\lambda| m$ on the algebra $A$ of finite disjoint unions of h-intervals, so they also agree on $\sigma(A)=\mathcal{B}_\mathbb{R}$ by Theorem 1.14. This means that \begin{equation}\tag{2} m(B+s)=m(B) \text{ and } m(\lambda B)=|\lambda| m(B) \text{ for every } B\in \mathcal{B}_\mathbb{R}. \end{equation} Applying (2) for the Borel null set $N$, we get \begin{equation}\tag{3} m(N+s)=0=m(N) \text{ and } m(\lambda N)=0=|\lambda| m(N). \end{equation}

Using the fact that $F\subseteq N$, we conclude that $F+s\subseteq N+s$ and $\lambda F\subseteq \lambda N$. Since m is a complete measure, we now use (3) to prove that $F+s,\lambda F$ are Lebesgue null sets.

Finally, using (2) once again for the Borel set $E$ and the fact that $F+s, F$ are Lebesgue null sets, we get for free that $$m(A+s)\stackrel{(1)}{=}m(E+s)+m(F+s)=m(E+s) =:m_s(E)\stackrel{(2)}{=}m(E)+0=m(E\cup F)=m(A)$$ and with the same argument we get $m(\lambda A)=|\lambda| m(A)$.

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  • $\begingroup$ (+1) Very nice first answer :) $\endgroup$
    – TheSimpliFire
    Jul 30, 2019 at 19:48
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    $\begingroup$ Thank you @TheSimpliFire, I hope it's correct too :) Any corrections/questions/remarks are appreciated! $\endgroup$ Jul 31, 2019 at 19:30
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Folland proves the theorem 1.21 in a different way, however the path you took is good took. Here is a proof following your path.

Let $L$ be the set of Lebesgue measurable sets in $\mathbb{R}$ and $m$ be the Lebesgue measure.

Theorem 1.21 - If $E\in L$ then $E+s\in L$ and $rE\in L$ for all $s,r\in\mathbb{R}$. Moreover, $m(E + s) = m(E)$ and $m(rE) = |r|m(E)$.

Proof:

First note that $\mathcal{I}$ the set of open open intervals is invariant by translations and dilatations. Moreover, if $I$ is an open interval then $m(I + s) = m(I)$ and $m(rI) = |r|m(I)$.

So,

for any set $C \subset \mathbb{R}$, we have $m^*(C + s) = m^*(C)$ and $m^*(rC) = |r|m^*(C)$

Now, suppose $E\in L$, then for any $Y \subset \mathbb{R}$, we have $$m^*(Y) = m^*(Y\cap E) + m^*(Y\cap E^c)$$

In particular,
$$m^*(Y-r) = m^*((Y-r)\cap E) + m^*((Y-r)\cap E^c)$$

So, since $E^c+r=(E+r)^c$, we get, \begin{align*} m^*(Y)= m^*(Y-r) &= m^*((Y-r)\cap E) + m^*((Y-r)\cap E^c)= \\ & = m^*(Y\cap (E+r)) + m^*(Y\cap (E^c+r))= \\ & = m^*(Y\cap (E+r)) + m^*(Y\cap (E+r)^c) \end{align*}

So $E+s \in L$, and since $m^*(E + s) = m^*(E)$,we have that $m(E + s) = m(E)$.

Now suppose $r\neq 0$. Then in a similar way, have

$$m^*(r^{-1}Y) = m^*((r^{-1}Y)\cap E) + m^*((r^{-1}Y)\cap E^c)$$

So, since $r(E^c)=(rE)^c$, we get, \begin{align*} m^*(Y)= |r|m^*(r^{-1}Y) &= |r|m^*((r^{-1}Y)\cap E) + |r|m^*((r^{-1}Y)\cap E^c)= \\ & = m^*(Y\cap (rE)) + m^*(Y\cap r(E^c))= \\ & = m^*(Y\cap (rE)) + m^*(Y\cap r(E^c)) \end{align*}

So $rE \in L$, and since $m^*(rE) = |r|m^*(E)$, we have that $m(rE) = |r|m(E)$.

Now, if $r=0$, then $rE=\emptyset$ (if $E=\emptyset$) or $rE=\{0\}$. In both cases it is trivial that $m(rE) = 0 = |r|m(E)$.

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  • $\begingroup$ $$|r|m^{*}((r^{-1}Y)\cap E) + |r|m^{*}((r^{-1}Y)\cap E^c)$$ in the next two steps where did the $|r|$ go? $\endgroup$
    – Wolfy
    Jun 16, 2016 at 20:58
  • $\begingroup$ @Wolfy Note that $$ (r^{-1}Y)\cap E =r^{-1}(Y \cap rE) $$ So, we have $$ |r|m^*((r^{-1}Y)\cap E)= |r|m^*(r^{-1}(Y \cap rE))=|r||r^{-1}| m^*(Y \cap rE)=m^*(Y \cap rE)$$ A similar argument applies to $(r^{-1}Y)\cap E^c$. $\endgroup$
    – Ramiro
    Jun 16, 2016 at 21:24
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"since $x\in E$ and $E\in L$ then $x\in L$" is gibberish. Individual real numbers are not members of $L.$ Go directly to the def'ns of $L$ and $m.$ See what you get if every set $S$ of reals mentioned in the def'ns is replaced by $x+S$ or if every $S$ is replaced by $rS.$

As someone said on this site, definitions are your friends. Call on them when you're stuck.

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