Show that if A is measurable set then the set 3A : = { 3a : a Є A} is measurable. Find its measure. How about the set -3A? Give proofs to your answers. [20 marks]
I am not sure how you are defining "measurable", I will assume you mean Jordan measurable.
Also where is $\displaystyle A$ a subset of? It will be assumed that $\displaystyle \emptyset \not = A\subset \mathbb{R}$.
Let $\displaystyle x_0 \in \partial A$ show that $\displaystyle 3x_0 \in \partial (3A)$ and conversely.
Let $\displaystyle \epsilon > 0$.
Since $\displaystyle A$ is measurable it means $\displaystyle \partial A$ can be covered by intervals $\displaystyle I_1, ... ,I_k$ so that $\displaystyle \Sigma_{i=1}^k \text{length}(I_i) < \epsilon/3$.
If $\displaystyle I_i = [a_i,b_i]$ define $\displaystyle J_i = [3a_i,3b_i]$.
Then $\displaystyle J_1,...,J_k$ cover $\displaystyle 3A$ and $\displaystyle \Sigma_{i=1}^k \text{length}(J_i) < \epsilon$.
Thus, $\displaystyle 3A$ is measurable.