# Sensitivity of Lebesgue measure of level sets

Assume we are given some constant $$\alpha$$ and a subset $$\Omega\subset\mathbb{R}^n$$ such that $$\lambda(\Omega)<\infty$$, where lambda denotes the Lebesgue measure. We consider the mapping $$\Lambda:L^1(\Omega)\to\mathbb{R}$$ with $$\Lambda(f)=\lambda(\{x\in\Omega: f(x)>\alpha\}).$$

I am trying to figure out a way to control the error sensitivity of this problem. If the function values are close to $$\alpha$$, then a small perturbation of the function could dramatically change the measure. Any suggestions as to what tools I could use would be greatly appreciated.

Thanks

• More precisely, you want to control the difference $$|\Lambda(f)-\Lambda(g)|,$$ right? – Giuseppe Negro Oct 29 '18 at 13:22
• Yes, exactly :) – Nicolas Bourbaki Oct 29 '18 at 13:25
• You could try to use Chebyshev's inequality en.wikipedia.org/wiki/… – daw Oct 29 '18 at 16:36

The following standard trick might be useful. Let $$\Lambda_\alpha(f):=|\{f>\alpha\}|.$$ Consider another function $$g\in L^1$$. If $$f(x)+g(x)>\alpha$$, then either $$f(x)$$ or $$g(x)$$ or both must be bigger than $$\alpha/2$$; $$\{f+g>\alpha\}\subset \{f>\tfrac\alpha2\}\cup\{g>\tfrac\alpha2\}.$$ Therefore $$\tag{1} \Lambda_\alpha(f+g)\le \Lambda_{\alpha/2}(f)+\Lambda_{\alpha/2}(g).$$ Now, adding and subtracting $$g$$, we see that $$\Lambda_{2\alpha}(f)\le \Lambda_{\alpha}(f-g) +\Lambda_\alpha(g),$$ which gives the bound $$\tag{2} \Lambda_{2\alpha}(f)-\Lambda_\alpha (g)\le \Lambda_{\alpha}(f-g).$$
This is not very pretty, but maybe it helps. If $$\alpha>0$$, then you can combine it with Chebyshev's inequality, yielding $$\max(\Lambda_{2\alpha}(f)-\Lambda_\alpha(g), \Lambda_{2\alpha}(g)-\Lambda_\alpha(f))\le \frac{1}{\alpha}\|f-g\|_1.$$