Let $ (\Omega, \Sigma, \mu) $ be a measure space and let $f$ be a real valued function on $\Omega $ such that $\mu (x :f(x)
Then the minimization problem I= $\inf_{g \in C} \int f(x)g(x) \mu(dx)$ is solved by $g(x)= \chi_{(f where $s=\sup(t: \mu((x :f(x)
My question is, how can one show that g(x) given above is the minimizer.
Thanks