Assume that $\lambda(A+t-A)=0$ for all $t \in Q$ show that either $A$ is a null set, or full measure.

Question

Assume that $\lambda(A+t-A)=0$ for all $t \in \mathbb{Q}$ show that either $A$ is a null set, or full measure. We will assume that $\lambda(A)>0$. I think I have a solution, but I would like to be sure it works, so i hope someone could check my work and perhaps provide alternative solution. Since $\lambda(A)>0$ we know that at least one point in $A$ is a lebesgue point, call it $x$. $$\lim_{\delta \to 0}\frac{\lambda(A\cap(x-\delta,x+\delta))}{2\delta}=1$$. Since $\mathbb{Q}$ is dense, there is $x_n \to x$, $x_n \in \mathbb{Q}$ so consider $$\frac{\lambda(A\cap(x_n-\delta,x_n+\delta))}{2\delta}=\frac{\lambda(A-x_n\cap(-\delta,\delta))}{2\delta}= \frac{\lambda(A\cap(-\delta,\delta))}{2\delta}$$ where the last equality comes from assumption. Now by continuity we have taking the limit $n\to \infty$ $$\frac{\lambda(A\cap(x-\delta,x+\delta))}{2\delta}= \frac{\lambda(A\cap(-\delta,\delta))}{2\delta}$$ now taking the limit we have that $\lim_{\delta \to 0}\frac{\lambda(A\cap(-\delta,\delta))}{2\delta}=1$ Now let $y \in \mathbb{R}$ and let $y_n \to y$ where $y_n \in \mathbb{Q}$ we repeat the argument and get that $$\lim_{\delta \to 0}\frac{\lambda(A\cap(y-\delta,y+\delta))}{2\delta}= \lim_{\delta \to 0}\frac{\lambda(A\cap(-\delta,\delta))}{2\delta}=1$$ thus all points in $\mathbb{R}$ are lebesgue points and so $A$ must be full measure. Is this correct?