Can some transformation of a function always be found such that it has a specified integral?

Question

Say that I've got a measurable function $\mathcal L({\bf x})$ that takes in some values ${\bf x} \in \mathbb R^{k}$ and returns a real number. $\mathcal L$ is bounded from above.

Can I always find a functional $f$ which transforms my function $\mathcal L$ such that, in some sense monotonicity is preserved - if ${\bf x}$ maximizes $\mathcal L$, it should also maximize $f(\mathcal L({\bf x}))$. Moreover, $f(\mathcal L({\bf x})) \geq 0$ and:

$$\int_{...}^{\infty} \int_{-\infty}^{\infty} f(\mathcal L({\bf x})) dx_1...dx_n = 1 \;?$$

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Thanks to everyone who took time to answer my question.

Answer 1

This is not possible without further assumptions on $\mathcal L$: Just consider the constant function $\mathcal L \equiv 1$. Then, for all $f$, your integral is either $0$ or $+\infty$.

Answer 2

If $\mathcal L$ can take negative values, then $f(0)>0$ and the integral will certainly diverge. Even with $f(\le0)=0$ the integral can diverge.