Proving that there's no translation invariant measure on the power set of $\mathbb{R}$ The goal of this task given to me is to show that there is no (non-trivial) translation invariant measure on $P(\mathbb{R})$, the power set of $\mathbb{R}$, and I think I almost completed it, but I just can't find a way to prove the very last bit that's missing. But let's start at the beginning.
Let $\mu: P(\mathbb{R}) \to [0, ∞]$ be a measure that satisfies the following conditions:
$(*) \mu([0, 1]) = 1$ and $\mu(x + A) = \mu(A)$ for all $x \in \mathbb{R}, A \subseteq \mathbb{R}$.
The assignment asks me to consider the equivalence relation 
$x \sim y :<=> x - y \in \mathbb{Q}$
Out of every equivalence class, we choose a representative $x \in [0, 1]$. Let $X \subseteq [0, 1]$ be the set of representatives that we got that way.
I am now to consider the sets $\mathbb{R} = \cup_{x \in X} (x + \mathbb{Q})$ and $\mathbb{R} = \cup_{q \in \mathbb{Q}} (X + q)$, and find a contradiction by showing that $\mu$ would need to satisfy both $\mu(X) = 0$ aswell as $\mu(X) > 0$.
What I've shown so far: I showed that $\mu(\mathbb{R}) = ∞$ (using the fact that $[0, 1]$ is sent to $1$), aswell as that $\mu(\{a\}) = 0$ for all $a \in \mathbb{R}$. From this, it also follows that all countable subsets of $\mathbb{R}$ are sent to $0$ by $\mu$ (so especially $\mu(\mathbb{Q}) = 0$).
I've also shown that, since $\mu(\cup_{q \in \mathbb{Q}} (X + q)) = ∞$ and since $\cup_{q \in \mathbb{Q}} (X + q)$ is a countable, disjoint union, we have that $\mu(X) > 0$.
The only part I'm still missing is to show that $\mu(X) = 0$ via the fact that $\mathbb{R} = \cup_{x \in X} (x + \mathbb{Q})$, as written above. I just can't get my head around how I could show this. $\cup_{x \in X} (x + \mathbb{Q})$ is a disjoint, but uncountable union because $X$ contains uncountably many elements; therefore, we can't use the $\sigma$-additivity of a measure. I see that $\mu(x + \mathbb{Q}) = 0$ for each $x \in X$, because $\mathbb{Q})$ is countable, but I don't know how that helps me.
I've also thought about decomposing $[0,1]$ into a disjoint union of sets, but that didn't lead anywhere so far. If we "remove" countably many points of $[0,1]$, the remaining set would still be sent to $1$ by $\mu$, for the reasons given above. So how could I conclude that $X$ must be sent to $0$? I'm out of ideas.
 A: I think that the question is not correct. If for each $Y \subset R$ we set $\mu(Y)=+\infty$ if $card(Y)>\aleph_0$ and $\mu(Y)=0,$ otherwise, then $\mu$ stands an example of nontrivial translation invariant measure in $R$ which vanishes on singletons.
If you require  that in addition  $\mu$ must be  $\sigma$-finite, then such stated  question is very old and it is not solvable  within the theory $ZF$. 
For example, if we  accept Axiom of Choice  then by using Ulam's  well known theorem, asserted that on the powerset of the $\aleph_1$ there does not exist a $\sigma$-finite measure  which vanishes on singletons, we get a negative answer.
But if we accept Steinhauss-Mycielski determinateness axiom about the existence of winning strategy then following Mycielski and Swierczkowskievery celebrated century theorem  asserted that under that axiom  each  subset of the real axis is Lebesgue measurable we can answer positively  on your question(cf. [Mycielski J., Swierczkowski S., On the Lebesgue measurability and the axiom of determinateness, Fund.,54,(1964),67-71]).
A: One way of doing this is in analogy with the construction of the Vitali set. 
Suppose $X$ is defined as in your question. Construct $\tilde{X}$ by taking each $x\in X$, and then translating it by a rational to the interval $[0,\epsilon)$ for some fixed $0<\epsilon<1$. Then the set $\tilde{X}$ is just $X$ where all representatives of the equivalence classes are chosen to lie in $[0,\epsilon)$. Arguing as you did, we can show $\mu(\tilde{X})>0$.
Now consider the set 
$$E=\cup_{q\in\mathbb{Q}\cap[0,1-\epsilon]}(\tilde{X}+q).$$
Then $E\subset [0,1]$, and so $\mu(E)\le 1$. On the other hand, as the sets $\tilde{X}+q$ are disjoint:
$$\mu(E)=\sum_{q\in \mathbb{Q}\cap[0,1-\epsilon]}\mu(\tilde{X}+q).$$
Applying translational invariance
$$=\sum_{q\in \mathbb{Q}\cap[0,1-\epsilon]}\mu(\tilde{X}).$$
The sum is bounded above by $1$, and has countably infinitely many terms of size $\mu(\tilde{X})$. We must therefore have $\mu(\tilde{X})=0$, which contradicts our earlier result that $\mu(\tilde{X})>0$.
