# Bathtub principle proof

Let $(\Omega, \Sigma, \mu)$ be a measure space and let $f$ be a real valued function on $\Omega$ such that $\mu (x :f(x)<t)$ is finite for all t $\in \mathbb{R}$. Let the number $G>0$ be and given and defined a class of measurable functions $\Omega$ by $C= (g: 0\le g(x) \le 1$ for all $x$ and $\int g(x)\mu(dx)=G)$

Then the minimization problem I= $\inf_{g \in C} \int f(x)g(x) \mu(dx)$ is solved by $g(x)= \chi_{(f<s)}(x) + c\chi_{(f=s)}(x)$ where $s=\sup(t: \mu((x :f(x)<t)) \le G)$ and $c\mu ((x :f(x)=s))=G-\mu ((x :f(x)<s))$

My question is, how can one show that g(x) given above is the minimizer.

Thanks

-
I think you have made a typo in your question.The $s$ should be $sup\{t:μ\{x:f(x)<t\}\le G\}$ and $μ(x:f(x)<t)$ is finite for all $t$. –  Ben Nov 24 '12 at 16:04
you are right, thanks Ben –  jake Nov 24 '12 at 17:30

## 1 Answer

Let $h$ be any other member of $C$. To show that $g$ is a minimizer, you need to establish that $I(g) \leq I(h)$, which is equivalent to showing $$\int f(g-h) \leq 0.$$

To show this, split the range of integration into the sub-level, sup-level, and level sets of $f$; i.e., $\{f<s\}$, $\{f>s\}$, and $\{f=s\}$:

\begin{align} \int f (g-h) &= \int\limits_{\{f<s\}} f (g-h) + \int\limits_{\{f>s\}} f (g - h) + \int\limits_{\{f=s\}} f (g - h) \\ &\leq s \int\limits_{\{f<s\}} (g-h) - \int\limits_{\{f > s\}} f h + \int\limits_{\{f=s\}} s (g-h) \\ &\leq s \int\limits_{\{f<s\}} (g-h) - s \int\limits_{\{f > s\}} h + s \int\limits_{\{f=s\}} (g-h) \\ & \leq s \left( \int\limits_{\{f<s\}} (g-h) + \int\limits_{\{f>s\}} (g-h) + \int\limits_{\{f=s\}} (g-h) \right) \\ & = s \int(g - h) = s (G -G) = 0. \end{align}

-
thanks rick, 15char –  jake Nov 28 '12 at 3:44