If $\lambda(A)=0$ and $\mu$ is $\sigma$-finite, then $\mu(A+x)=0$ for $x$ outside a set of Lebesgue measure $0$

Question

I have a measure associated with a distribution function (but not necessarily a probability distribution function) on $\mathbb{R}$, namely

$\mu((a,b\rbrack)=F(b)-F(a)$.

Now, $F$ is not absolutely continuous, and therefore there exists a Lebesgue set $A$ such that $\mu(A)>0$. I need to prove that if $\lambda(A)=0$ and $\mu$ is $\sigma$-finite, then $\mu(A+x)=0$ for $x$ outside a set of Lebesgue measure $0$.

I have a hint that says to use Fubini's theorem, $\lambda$'s invariance under translations and reflection through $0$.

What I thought was that since $\lambda(A)=0$, it is possible to find $B=\cup_n I_n$ such that $A\subseteq B$ and $\lambda(B)<\epsilon$. But then I could cover $\mathbb{R}$ using translates of $B$ (that I'll call $B_x$). If $C=\{x\colon \mu(B_x)>0 \}$ has nonzero measure, and $\mu(\mathbb{R})>\infty$, a contradiction.

Is this the right approach? I don't see how to use the hints. Could someone please point me in the right direction?

Answer 1

Try this: compute $$ \int_{\mathbb R} \mu(A + x) d\lambda(x) = \int_{\mathbb R} \int_{\mathbb R} \chi_{A + x}(y) \, d\mu(y) d\lambda(x).$$ Measurability issues aside, Tonelli's theorem implies that this equals $$\int_{\mathbb R} \int_{\mathbb R} \chi_{A +x}(y) d\lambda(x) d\mu(y).$$ Since $$\chi_{A + x}(y) = 1 \iff y \in A + x \iff x \in -A + y \iff \chi_{-A + y}(x) = 1$$ you get $$\int_{\mathbb R} \chi_{A +x}(y) d\lambda(x) = \int_{\mathbb R} \chi_{-A + y}(x) d\lambda(x) = \lambda(-A + y) = \lambda(-A) = \lambda(A) = 0.$$ Hence $$\int_{\mathbb R} \mu(A + x) d\lambda(x) = 0$$ so that $\mu(A + x) = 0$ for $\lambda$ almost every $x$.

Answer 2

Let $\phi(x,y) = 1_A(y-x)$.

We need to ensure that $\phi$ is $\mu \times \lambda$ measurable. The function $(x,y) \to x-y$ is measurable, hence the set $\{ (x,y) | x-y \in A \}$ is measurable, and so is the complement. Then the set $\{(x,y) | \phi(x,y) < \beta \}$ is either empty, the entire space of the complement of the set $\{ (x,y) | x-y \in A \}$ and so $\phi$ is measurable.

Note that $\mu$ is automatically $\sigma$-finite, since $\mu(-n,n]) < \infty$ for all $n$.

Tonelli gives $\int \phi d \mu \times \lambda =\int ( \int \phi(x,y) d \lambda(x) ) d \mu(y) = \int ( \int 1_{-A + \{y\}} (x) d \lambda(x) ) d \mu(y) = 0$, hence $\int \phi(x,y) d \mu(y) = \int 1_{A + \{x\}} (y) d \mu(y) = \mu (A + \{x\}) = 0$ for ae. [$\lambda$] $x$.