I am currently self teaching myself measure theory from https://terrytao.files.wordpress.com/2012/12/gsm-126-tao5-measure-book.pdf .
I would like some help/tips in proving the Boolean closure properties in Exercise 1.1.6. I am first proving that if $E$ and $F$ are Jordan measurable then $E\cap F$ is Jordan measurable, it seems very elementary and so I don't know why it is causing difficulty. Here is my attempt of a proof:
If $E$ and $F$ are Jordan measurable then there exists elementary sets $A_1,B_1,A_2,B_2$ such that $$ A_1\subset E\subset B_1, A_2 \subset F \subset B_2$$ with $m(A_1) = m(B_1), m(A_2) = m(B_2).$ Note that $$ A_1 \cap A_2 \subset E\cap F \subset B_1 \cap B_2.$$ I now want to show that $m(A_1 \cap A_2) = m(B_1\cap B_2)$ and then since $A_1 \cap A_2$ and $B_1 \cap B_2$ are elementary $E\cap F$ would be Jordan measurable. I also know that $m(B_1 \cap B_2) \leq m(A_1\cap A_2)$ by the monotonicity property of elementary sets. I was trying to partition each of the sets $A_1,B_1,A_2,B_2$ e.g. $$ A_1 = (A_1 \cap A_2)\cup (A_1\backslash A_2).$$ Using this we can conclude $$ m(A_1 \cap A_2) + m(A_1\backslash A_2) = m(B_1 \cap B_2) + m(B_1\backslash B_2) $$ $$ m(A_1 \cap A_2) + m(A_2\backslash A_1) = m(B_1 \cap B_2) + m(B_2\backslash B_1). $$ This doesn't seem to help and I've tried various other things to no avail. Thanks for your help.