A really strange question here in my opinion:

Let $A,B$ be two measurable subsets of ${\mathbb{R}}^{1}$. Define $f(x)=|(A-x)\cap B|$. Evaluate ${\int}_{{\mathbb{R}}^{1}}fdx$. Here $|\cdot |$ refers to the measure.

Here $f$ is clearly non-negative, so I tried using the definition of Lebesgue integral:

${\int}_{{\mathbb{R}}^{1}}fdx=sup\{{\int}_{{\mathbb{R}}^{1}}sdx:0\le s\le f\},$

where s are simple functions. But I realised that I couldn't come up with any simple functions.

I tried breaking up ${\mathbb{R}}^{1}$ into two parts:

${E}_{1}=\{y\in {\mathbb{R}}^{1}:y\in A\cap B\},{E}_{2}=\{y\in {\mathbb{R}}^{1}:y\notin A\cap B\}$

and then considering the sum

${\int}_{{E}_{1}}fdx+{\int}_{{E}_{2}}fdx$

and then I'm clueless as to how to proceed. I'm guessing the answer should be something intuitive like $|A\cap B|$. Any help would be appreciated.

Let $A,B$ be two measurable subsets of ${\mathbb{R}}^{1}$. Define $f(x)=|(A-x)\cap B|$. Evaluate ${\int}_{{\mathbb{R}}^{1}}fdx$. Here $|\cdot |$ refers to the measure.

Here $f$ is clearly non-negative, so I tried using the definition of Lebesgue integral:

${\int}_{{\mathbb{R}}^{1}}fdx=sup\{{\int}_{{\mathbb{R}}^{1}}sdx:0\le s\le f\},$

where s are simple functions. But I realised that I couldn't come up with any simple functions.

I tried breaking up ${\mathbb{R}}^{1}$ into two parts:

${E}_{1}=\{y\in {\mathbb{R}}^{1}:y\in A\cap B\},{E}_{2}=\{y\in {\mathbb{R}}^{1}:y\notin A\cap B\}$

and then considering the sum

${\int}_{{E}_{1}}fdx+{\int}_{{E}_{2}}fdx$

and then I'm clueless as to how to proceed. I'm guessing the answer should be something intuitive like $|A\cap B|$. Any help would be appreciated.