I am wondering about part 3 of the following question. I was hoping to get another hint. Thank you.
Identify the interval $[0,1]$ with the unit circle $S^1=\{e^{i\theta}:\theta\in[0,2\pi)\}\subset\mathbb C$. We will construct a set in $S^1$ that is not Borel-measurable. To this end, for $z=e^{i\alpha},w=e^{i\beta}\in S^1$, we say $z\sim w$ if $\alpha-\beta\in\mathbb{Q}$. This is clearly an equivalence relation. Using the axiom of choice we can construct a set $\Lambda$ whose elements are one from each equivalence class. For $\alpha\in[0,2\pi)\cap\mathbb{Q}$ let $\Lambda_\alpha=e^{i\alpha}\Lambda$ be the rotation of $\Lambda$ by the angle $\alpha$.
1) Prove that if $\alpha,\beta\in[0,2\pi)\cap\mathbb{Q}$ are distinct, then $\Lambda_\alpha\cap\Lambda_\beta=\varnothing$.
2) Prove that $S^1$ is the union over $\alpha\in[0,2\pi)\cap\mathbb{Q}$ of $\Lambda_\alpha$.
3) Let $m$ be the Lebesgue measure on $S^1$ (equipped with the Borel $\sigma$-algebra). Prove that $m(\Lambda)$ cannot exist. (Hint: use the thing you just proved and the fact that $m$ is shift-invariant.)
Some quick notes on the problem:
A relation of equivalence breaks the entire circle into disjoint equivalence classes. In this problem, we have countable many equivalence classes (ie. one from each rational number). In other words, to get an equivalence class, we pick one element from each class. To get an equivalence class, I pick an element, say $\gamma$ from $\Lambda$ and then we have $\{\gamma+\alpha: \alpha \in \mathbb{Q}\}$ is the set of all elements equivalent to $\gamma$.
As Henry hinted you, you should assume by contradiction that $m(\Lambda)=c > 0$.
Then you get $$\infty > m(S^1) = m(\bigcup_\alpha \Lambda_\alpha) = \sum_{\alpha} m(\Lambda_\alpha) = \sum_{\alpha} m(\Lambda) = \sum_{\alpha} c = \infty$$.
I would have loved commenting this as a comment instead but I don't have enough reputation yet.