# Show $\lvert\,f(x)-f(y)\,\rvert \leq 2\,\lvert\,x-y\,\rvert$ for Lebesgue measure

Having some trouble with Borel algebra understanding

My textbook says, let $$\lambda$$ be the Lebesgue measure on $$\mathbb{R}$$ and let $$A \in \mathcal{B}(\mathbb{R})$$. Define a function f: $$[0,\infty)\rightarrow \mathbb{R}$$ by f(x)=$$\lambda(A \cap [-x,x])$$

Show that $$\lvert\,f(x)-f(y)\,\rvert \leq 2\,\lvert\,x-y\,\rvert$$.

My ideas are:

1. $$|\lambda(A \cap [-x,x])-\lambda(A \cap [-y,y])|=|\lambda(A \cap \cap_{n=1}^\infty (-x-1/n,x+1/n))-\lambda(A \cap \cap_{n=1}^\infty (-y-1/n,y+1/n))|$$

2. $$|\lambda(A \cap [-x,x])-\lambda(A \cap [-y,y]|=|\lambda((-x,x))-\lambda(A \cap (-y,y))|$$

Not sure if either are correct. It says that the Boreal algebra is the collection of all open sets (equivalent closed sets).

But does this mean that I should write $$[-x,x]$$ as a union of open sets (1) so that they can be joined with A. Or does it mean that $$[-x,x]$$ becomes $$(-x,x)$$ (2) because there are only open sets in the Borel algebra

• I hope your textbook doesn't really say "that the Boreal algebra is the collection of all open sets" --- and not just because of the misspelling of "Borel". – Andreas Blass Aug 14 at 14:01
• – Martin R Aug 14 at 14:01
• @AndreasBlass No, that was my phrase – Daniel Aug 14 at 14:33
• Made an edit, forgot to add the $\mathcal{B}$ :-( – Daniel Aug 14 at 14:35

If $$x>y$$, then $$f(x)-f(y)=\lambda((A\cap[-x,-y))\cup(A\cap(y,x])) \leq\lambda([-x,-y)\cup(y,x])=\lambda([-x,-y))+\lambda((y,x])=2(x-y)$$ and the argument for $$x is similar with the roles of $$x$$ and $$y$$ reversed.
I suppose that $$A$$ should be Borel-measurable, should it? Otherwise I don't see where the Borel-Sigma-Algebra comes into place in you exercise. But it is not needed for the proof. You just have to assume that $$A$$ is Lebesgue-measurable in order to have the measure of it defined.
To prove your estimation you can assume wlog that $$y > x$$ and then $$A \cap [-x,x] \subseteq A \cap [-y,y]$$ and therefore $$|f(y)-f(x)| = \lambda(A \cap [-y,y]) - \lambda(A \cap [-x,x]) = \lambda((A \cap [-y,y]) \setminus (A \cap [-x,x])) = \lambda((A \cap [-y,-x]) \cup (A \cap [x,y])) \leq \lambda([-y,-x] \cup [x,y]) = 2|y-x|$$