Assume $f=g~~a.e$ with respect to a Lebesgue measure what is $L_g?$

Question

Suppose $f$ and $g$ are in $L^1(\mathbb{R}^n,m)$ where $m$ is a Lebesgue measure. And suppose that $f=g$ almost everywhere with respect to $m$.

We denote by $$L_f=\left\{x: \frac{1}{m(B(r,x))}\int_{B(r,x)}|f(x)-f(y)| \; dm(y) = 0\right\}$$ where $m$ is a Lebesgue measure.

Can $L_f$ be described in term of $L_g$?

Thanks for your help

Answer 1

Let $A$ be a subset such that $m(A)=0$ and $f(x)=g(x)~~~x\in \Bbb R^n\setminus A$

Then for $x\in \Bbb R^n\setminus A$ if $x\in L_f$ then we have

$$f(x)-f(y)=g(x)-g(y) ~~~\mbox{for all most every $y$}$$ that is $$\frac{1}{m(B(r,x))}\int_{B(r,x)}|g(x)-g(y)|\;dm(y) = \frac{1}{m(B(r,x))}\int_{B(r,x)}|f(x)-f(y)|\;dm(y)=0$$

Hence $x\in L_g$ therefore there exists $A$ such that $m(A)=0$ and $$L_g\cap\Bbb R^n\setminus A =L_f\cap\Bbb R^n\setminus A$$

Answer 2

Let $A=\{x:f(x)=g(x)\}$. Then, $m(A^c)=0$. Note that $$m(L_f) = m(L_f\cap A)+m(L_f\cap A^c)=m(L_f\cap A)$$ But $L_f\cap A=L_g\cap A$. I left to you the details.