Show that if $A,B$ are measurable, $A\subset E\subset B$, and $m(A)=m(B)$, then $E$ is measurable.

Question

Here's the full problem:

Suppose $A\subset E\subset B$, where $A,B$ are measurable with finite measure. Show that if $m(A)=m(B)$, then $E$ is measurable.

Here, we are dealing with measure space $(\mathbb{R},\mathcal{M},m)$. I come here just to see if this is a valid proof. Here we go:

Proof

We have that $m(A),m(B)<\infty$. Supposing $m(A)=m(B)$, we have

$$m(A)\leq m(E) \leq m(B)=m(A)$$ $$\implies m(A)=m(E)$$

Since $A$ is measurable, it follows that $E$ must then be measruable.

I know of another way to prove this, I was just wondering if this was a valid argument as well.

Answer 1

You assumed that $E$ is measurable when you wrote $$ m(A) \le m(E) \le m(B) $$ It's clear that if $m(E)$ is defined, then it must be equal to the common value of $m(A)$ and $m(B)$. The issue is how we know it's defined at all.

Answer 2

This is true if the measure involved is complete.

Answer 3

You have got to be careful! you cannot write $m(E)$ before knowing that it is measurable! :)

This is how I would prove the result:

By assumption $m(B) - m(A) = 0$, hence $m(B \setminus A) = 0$. To make this precise I need to use that $A \subset B$: $$ 0 = m(B) - m(A) = m(A) + m(B \setminus A) - m(A) = m(B\setminus A).$$

Now that we have this, denote by $m^*$ the outer measure (which is defined on every set, hence we can make sense of $m^*(E)$!). Recall also that for measurable sets $m^* = m$.

To prove that $E$ is measurable I am going to write it as the union of two measurable sets: $A$ and $E\setminus A$. Of course we don't know yet that the latter is measurable, but follows from the previous computation and the completeness of the measure (which I guess I can assume): $$m^*(E \setminus A) \le m^*(B \setminus A) = m(B \setminus A) = 0.$$

Hopefully I got it right :D