A question on excess mass

Question

The definition of the $\lambda-$excess mass of a measurable set $G$ in $R^N$ is as follows:

$$M(G)=\int_{G}fdx-\lambda\text{mes}(G)$$

where $\text{mes}(G)$ denotes the Lebesgue measure of $G$ and $f$ is a denisty function. Through some examples, I have noticed that $M(G_f(\lambda))$ where $G_f(\lambda)=\{x:f(x)\geq \lambda\}$ is greater than $\lambda-$excess mass of any other $G$. I don't know if this is correct or not.

Answer 1

This is true: Fix $\lambda$, $f$, and set $H:= G_f(\lambda)$.

For any $G$ we have $$ \int_G f\, dx = \int_{G\cap H} f\, dx + \int_{G\setminus H} f\, dx\leq \int_{G\cap H} f\, dx + \lambda \text{mes}(G\setminus H). $$ Therefore $$ \int_G f\, dx -\lambda\text{mes}(G)\leq \int_{G\cap H}f\, dx - \lambda\text{mes}(G\cap H). $$ Now do the same in reverse to add back the portion of $H$ missing: \begin{align} \int_{G\cap H}f\, dx -\lambda\text{mes}(G\cap H) &= \int_{G\cap H} f\, dx + \lambda\text{mes}(H\setminus G) -\lambda \text{mes}(H)\\ & \leq \int_H f\, dx -\lambda \text{mes}(H). \end{align}