Exercise on the properties of Lebesgue measurable sets Hey I have problems with this exercise.
We consider the set of rectangles in $\mathbb{R}^2$ $M := \left \{I_1 × I_2 : I_1, I_2 ⊆ \mathbb{R} \text{ intervals}\right \}$,
and a function $µ: M → [0, \infty]$ that satisfies the following conditions:
(1) Translation invariance: For all $a \in \mathbb{R}^2$ and all $Q ∈ M$, $µ(a + Q) = µ(Q)$
(2) Normalization: $µ([0, 1] × [0, 1]) = 1$
(3) Finite additivity: If a rectangle $Q$ is the union of two disjoint
rectangles $Q_1$ and $Q_2$, then $µ(Q) = µ(Q_1) + µ(Q_2)$
I have to prove the following statements:
a) The function $µ$ is monotone, that means, for all $Q_1, Q_2 \in M$ with $Q_1 ⊆ Q_2$ follows $µ(Q_1) ≤ µ(Q_2)$.
b) For all $x \in \mathbb{R}$ we have $µ(\left \{ x \right \}×[0, 1]) = 0$.
c) For $f : R^+\rightarrow R$, $f(x) := µ([0, x) × [0, 1])$ it holds that $f(x) = x$.
Hint: First consider $x \in  \mathbb{Q}$.
d) For all $a_1 ≤ b_1$ and $a_2 ≤ b_2$ we have $µ([a_1, b_1] × [a_2, b_2])= (b_1 − a_1) · (b_2 − a_2)$.
What I have done is the following.
I have proven a)
for b) I have some difficulties: What I have done is the following:
Since there is no $Q=[a,b]×[c,d]⊆ \left \{ x \right \}×[0, 1]$ than it follows that $µ(\left \{ x \right \}×[0, 1]) = 0$.
However, I don't think this is enough to confirm it.
for c) and d) I don't have any idea.
Can someone help me?
 A: I would do b) by contradiction. Assume $\mu(\lbrace x\rbrace\times [0,1])$ is non-zero, and show it cannot satisfy normalisation. Indeed, this is not hard using that $[0,1]$ has infinitely many points, together with finite additivity. Ask if you want more of a sketch.
EDIT: By request, further details. Let assume for contradiction that $\mu(\lbrace x\rbrace\times [0,1])=\epsilon>0$. Then let $N:=\lceil\frac1\epsilon\rceil+1\in\mathbb{N}$, so that $N\cdot\epsilon >1$. Then we just need $N$ points in $[0,1]$ to get our contradiction - pick for example $S=\lbrace\frac1N,\frac2N,\dots,\frac{N-1}{N},1\rbrace$. Then $S\subset [0,1]$, so we clearly have $\mu(S\times[0,1])\leq \mu([0,1]\times[0,1])=1$ by monotonicity, but using finite additivity, we can calculate
$$\mu(S\times[0,1]) = \sum_{i=1}^N\mu\left(\lbrace \frac{i}{N}\rbrace \times [0,1]\right) = N\cdot \mu\left(\lbrace 0\rbrace\times[0,1]\right)=N\cdot \epsilon>1.$$
Here the penultimate inequality follows from translation invariance. This is a contradiction.
EDIT END.
Importantly, note that b) implies (*) $\mu([a,b]\times[0,1])=\mu((a,b]\times[0,1]) $ by writing $[a,b]=\lbrace a\rbrace\cup (a,b]$ and using finite additivity.
Now, for c), by monotonicity it is clear that $f$ is increasing (make a small argument). Also, $f$ is additive, i.e. $f(x+y)=f(x)+f(y)$ (again, a small argument using finite additivity).
Furthermore, you can determine $f(\frac1n)$ for any $n\in \mathbb{N}$ as follows:
Use that
$$[0,1] = [0,\frac1n]\cup(\frac1n,\frac2n]\cup\dots\cup (\frac{n-1}{n},1]$$
to write
$$1= \mu([0,1]\times[0,1]) = \mu([0,\frac1n]\times[0,1]\cup(\frac1n,\frac2n]\times[0,1]\cup\dots\cup (\frac{n-1}{n},1]\times[0,1])$$
$$=\mu([0,\frac1n]\times[0,1])+\dots+\mu((0,\frac1n]\times[0,1]) = n\cdot \mu([0,\frac1n]\times[0,1]) = n\cdot f(\frac1n),$$
so you get that $f(\frac1n)=\frac1n$. Here, we have used finite additivity, translation invariance and the property (*) from b). Using that $f$ is additive, you now know $f(q)=q$ for all $q\in\mathbb{Q}$. As $\mathbb{Q}$ is dense in $\mathbb{R}$, you can determine $f$ on all of $\mathbb{R}$ using that $f$ is increasing.
Finally d): By complete symmetry of our argument in c), we see that $g(y):= \mu([0,1]\times[0,y]) = y$. In fact, a tiny argument gives you that we have the function $h(x,y) := \mu([0,x]\times[0,y])$ satisfies $h(x,y)=xy$. Now you just need translation invariance, and you are done.
EDIT: Here is the translation invariance argument. Given $[a_1,b_1]\times[a_2,b_2]$, let $a$ be the point $a = (-a_1,-a_2)$. Then
$$\mu([a_1,b_1]\times[a_2,b_2]) = \mu(a+[a_1,b_1]\times[a_2,b_2]) $$
$$= \mu([0,b_1-a_1]\times[0,b_2-a_2]) = (b_1-a_1)(b_2-a_2).$$
