The Lebesgue Measurability of Translates of Lebesgue Measurable Sets

Recall from Translation Invariance of the Lebesgue Outer Measure page that if $E \in \mathcal P (\mathbb{R})$ and $a \in \mathbb{R}$ then:

(1)

\begin{align} \quad m^*(E) = m^*(E + a) \end{align}

Where $E + a = \{ e + a : e \in E \}$. We now prove that if $E$ is a Lebesgue measurable set then so is $E + a$.

Theorem 1: If $E \in \mathcal P(\mathbb{R})$ is a Lebesgue measurable set and $a \in \mathbb{R}$ then $E + a$ is a Lebesgue measurable set.

• Proof: Let $E \in \mathcal P(\mathbb{R})$ be a Lebesgue measurable set and let $A \in \mathcal P(\mathbb{R})$. Then:

(2)

\begin{align} \quad m^*(A \cap (E + a)) + m^*(A \cap (E + a)^c) = m^*((A - a) \cap E + a) + m^*((A - a) \cap E^c + a) \end{align}

• By the translation invariance of the Lebesgue outer measure we have that:

(3)

\begin{align} \quad m^*(A \cap (E + a)) + m^*(A \cap (E + a)^c) = m^*((A - a) \cap E) + m^*((A - a) \cap E^c) \end{align}

• And since $E$ is Lebesgue measurable we have that:

(4)

\begin{align} \quad m^*(A \cap (E + a)) + m^*(A \cap (E+a)^c) = m^*((A - a)) = m^*(A) \end{align}

• Therefore $m^*(A) = m^*(A \cap (E + a)) + m^*(A \cap (E+a)^c)$, so $E + a$ is Lebesgue measurable. $\blacksquare$