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.


3 Answers 3


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

  • $\begingroup$ (+1) Very nice first answer :) $\endgroup$
    – TheSimpliFire
    Jul 30, 2019 at 19:48
  • 1
    $\begingroup$ Thank you @TheSimpliFire, I hope it's correct too :) Any corrections/questions/remarks are appreciated! $\endgroup$ 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 $m(rE) = |r|m(E)$.


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 $m(rI) = |r|m(I)$.


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

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

"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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .