Show that if $A,B$ are measurable, $A\subset E\subset B$, and $m(A)=m(B)$, then $E$ is measurable. 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.
 A: 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
A: 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.
A: This is true if the measure involved is complete.
