# Real Analysis, Folland Theorem 1.21 Borel Measures

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.

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 $$$$\tag{1} A+s=(E\cup F)+s=(E+s)\cup(F+s) \text{ and } \lambda A=\lambda E \cup \lambda F$$$$ we need to prove that $$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 $$$$\tag{2} m(B+s)=m(B) \text{ and } m(\lambda B)=|\lambda| m(B) \text{ for every } B\in \mathcal{B}_\mathbb{R}.$$$$ Applying (2) for the Borel null set $$N$$, we get $$$$\tag{3} m(N+s)=0=m(N) \text{ and } m(\lambda N)=0=|\lambda| m(N).$$$$

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 $$m(\lambda A)=|\lambda| m(A)$$.

• (+1) Very nice first answer :) Jul 30, 2019 at 19:48
• Thank you @TheSimpliFire, I hope it's correct too :) Any corrections/questions/remarks are appreciated! Jul 31, 2019 at 19:30

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 $m(rE) = 0 = |r|m(E)$.

• $$|r|m^{*}((r^{-1}Y)\cap E) + |r|m^{*}((r^{-1}Y)\cap E^c)$$ in the next two steps where did the $|r|$ go? Jun 16, 2016 at 20:58
• @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 $(r^{-1}Y)\cap E^c$. Jun 16, 2016 at 21:24

"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.