Help verify a solution showing $f\left(x \right)=\int_\Bbb{R} {{\chi _A}\left(y \right){\chi _B}\left( {x-y} \right)dy} $ is well-defined everywhere

Question

The question is,

Let $A,B⊂[0,1]$ be measurable sets with $|A|>1/2$,$|B|>1/2$ where $|*|$ denotes Lebesgue measure. Prove that

a. $|A⋂(1-B)|>0$ where $1-B≔{1-x:x∈B}$ and conclude that there exist $x∈A$,$y∈B$ st. $x+y=1$.

b. the function $f\left( x \right) = \int_\Bbb{R} {{\chi _A}\left( y \right){\chi _B}\left( {x - y} \right)dy} $ is well-defined, i.e. $f(x)<+∞$, for all $x∈R$.

The following is my solution. I think my solution should be correct, but I want to make sure, especially for b), for which I am not very confident.

Thank you!

Answer 1

Your solution is correct for (a). For (b), it is better like \begin{align} f\left( x \right) &= \int_\Bbb{R} {{\chi _A}\left( y \right){\chi _B}\left( {x - y} \right)dy} \\ &= \int_\mathbb{R} {{\chi _A}\left( y \right){\chi _{x-B}}\left( {y} \right)dy} \\ &= \int_{A\cap(x-B)} {{\chi(y)}\:dy} \\ &\leqslant m(A\cap(x-B)) \\ &<\infty \end{align}