# Question on maximization over functions

Suppose I want to solve the following problem

$$\max_{f(\cdot)} \int f(x) g(x) d\mu(x)$$

where the maximization is over measurable functions $$f:X\to [0,1]$$ , $$\mu$$ is a finite measure and $$g$$ is measurable.

Can someone come up with an example such that the problem does NOT have a solution? I imagine one can try to maximize point by point and obtain something that is not measurable and then the problem would not have a solution?

• If you maximize over all measurable functions you will always get $\infty$ as long as $\int g d\mu$ is not zero. Aug 9, 2019 at 23:40
• Thanks. I forgot to write that $f$ goes to $[0,1]$. Fixed it. Aug 9, 2019 at 23:42
• I think you still need some condition on $g$. For example, take $\mu$ to be the Lebesgue measure restricted to the interval (0, 1] and $g(x)=1/x$ for $x > 0$. Then $\int g d\mu = \int_0^1 g(x) dx = +\infty$. Now take $f(x)=1$. Aug 10, 2019 at 2:19
• Maybe $g$ integrable is missing? Then $\int f(x) g(x) d\mu(x) = \int f(x) g^+(x) d\mu(x) - \int f(x) g^-(x) d\mu(x) \leq \int g^+(x) d\mu(x)$. And the maximum is therefore reached for $f = 1_{X^+}$, where $X^+ = \{x \in X \ s.t. \ g(x) > 0 \}$ Aug 10, 2019 at 2:31
• I'm looking for conditions for there not be a solution. Can anyone think of an example? Aug 11, 2019 at 19:08

If $$g$$ is not integrable the problem can be unbounded: Let $$\mu$$ be the Cauchy distribution on the real line. Let $$g(x) = |x|$$. Then, we can take $$f(x) = 1$$ for all $$x$$ and we have $$\int f g d\mu = \int |x| d\mu(x) = \infty$$.