Prove that $A$ is Lebesgue measurable implies that $x+A = \{ x+y : y \in A \}$ is measurable This question comes from Exercise 4.5 of Real Analysis for Graduate Students by Bass. After some deduction I reduced the question to the following form:
Show that if $A$ is a Lebesgue measurable set on $\mathbb R$, then $x+A = \{ x+y : y \in A \}$ and $cA = \{ cy : y \in A \} $, $c \in \mathbb R$ are both measurable.
While it's tempting to take it for granted, I find it difficult to prove it by directly using the definition of Lebesgue measure, because it's not directly constructed from the countable union and complement of Lebesgue measurable sets. Can anyone give a hint? 
 A: Claim 1:  $A\subset \mathbb{R}$ is Lebesgue measurable if and only if $x + A$ is Lebesgue measurable.
Proof of claim1:  It is enough to show only one direction, since $A = \big( x + A \big) + (-x).$ Suppose $A$ is measurable, then for any $B\subset \mathbb{R}$, we have 
 $$m\big( B \big) \geq m\big( B \cap A \big) + m\big( B \cap A^c \big) \,\, . $$
Therefore, 
$ m\big( x + B \big) \geq  m\big( B \cap A \big) + m\big( B \cap A^c \big) = m\Big( (B + x) \cap (A + x) \Big) + m\Big( (B + x) \cap (x + A^c) \Big)$,
which implies that $x + A$ is Lebesgue measurable.  ___   Q.E.D.
Claim 2:   $m(cA) = \vert c \vert\cdot m(A)$ for any $A\subset \mathbb{R}$ and any $c\in\mathbb{R}$.  The case where $c =0$ is trivial. Now without losing any generality, we assume that $c > 0$. Then,
\begin{align*}
c\cdot m\big(  A \big) &= c\cdot \inf\left\{ \,\,  \sum_{n=1}^\infty \text{diam}I_n \,\, : \,\,  A \subset \bigcup_{n=1}^\infty I_n\,\,,\quad\text{where $I_n$'s are bounded, mutually disjoint intervals} \,\, \right\} \\
& =  \inf\left\{ \,\,  c\cdot \sum_{n=1}^\infty  \text{diam} I_n \,\, : \,\,  A \subset \bigcup_{n=1}^\infty I_n\,\,\, \right\} \\
& = \inf\left\{ \,\,   \sum_{n=1}^\infty  \text{diam}\big(c I_n \big) \,\, : \,\,  A \subset \bigcup_{n=1}^\infty I_n\,\,\, \right\} \\
& = \inf\left\{ \,\,   \sum_{n=1}^\infty  \text{diam}\big(c I_n \big) \,\, : \,\,  cA \subset \bigcup_{n=1}^\infty cI_n\,\,\, \right\}   \\
& = m\big( cA \big) \,\, .
\end{align*}
Similarly, one can also show that for $c > 0$, $A$ is measurable if and only if $cA$ is measurable.  ___Q.E.D.
