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
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 $$ \max(\Lambda_{2\alpha}(f)-\Lambda_\alpha(g), \Lambda_{2\alpha}(g)-\Lambda_\alpha(f))\le \frac{1}{\alpha}\|f-g\|_1.$$