# Intersection of Jordan measurable sets is Jordan measurable

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.

First of all, note that $E$ being Jordan measurable doesn't necessarily implies $m(A_1)=m(B_1)$ in your question. It means, by definition, that $\underset{A\subset E}\sup m(A)=\underset{B\supset E}\inf m(B)$, with $A,B:$ elementary.
To answer your qeustion, I suggest you first prove a simple and useful lemma, which is exactly Tao's Exercise 1.1.5, i.e. $E$ is Jordan measurable iff $\forall \epsilon > 0$, there exists elementary sets $A\subset E\subset B$ such that $m(B\setminus A) \le \epsilon$.
Now we can prove the claim as follows. $\forall \epsilon>0$, there exist elementary sets $A_1\subset E\subset B_1$ and $A_2\subset F\subset B_2$ with $m(B_1\setminus A_1)\le \epsilon/2$ and $m(B_2\setminus A_2)\le \epsilon/2$. As you noted, we have $(A_1\cap A_2)\subset (E\cap F)\subset (B_1\cap B_2)$. Letting $B\triangleq B_1\cap B_2$, the key is only to convince yourself that $B \setminus(A_1\cap A_2)= (B\setminus A_1)\cup (B\setminus A_1)\subset (B_1\setminus A_1)\cup(B_2\setminus A_2).$ The conclusion then follows from the subadditivity of the elementary measure, i.e. $$m((B_1\cap B_2) \setminus(A_1\cap A_2))\le m(B_1 \setminus A_1)+m(B_2 \setminus A_2)\le \epsilon.$$