# In an inequality of integrals, can I multiply both integrands with the same non-negative, monotonous function?

Assume that you have two functions $f^-,f^+ :(\mathbb R^+)^n \rightarrow \mathbb R$. Also assume that every integral I use converges. Assume that we have the following inequality for every $(a_1,...,a_n) \in (\mathbb R^+)^n$: \begin{align*} \int\limits_{a_1}^{\infty}...\int\limits_{a_n}^{\infty} f^-(x_1,...,x_n)dx_1...dx_n \leq \int\limits_{a_1}^{\infty}...\int\limits_{a_n}^{\infty} f^+(x_1,...,x_n)dx_1...dx_n \end{align*}

Now suppose that we have a function $h:(\mathbb R^+)^n \rightarrow \mathbb R^+$ which is monotonously increasing in every component. Does then the following hold:

\begin{align*} \int\limits_{a_1}^{\infty}...\int\limits_{a_n}^{\infty} h(x_1,...,x_n) \cdot f^-(x_1,...,x_n)dx_1...dx_n \leq \\ \int\limits_{a_1}^{\infty}...\int\limits_{a_n}^{\infty} h(x_1,...,x_n) \cdot f^+(x_1,...,x_n)dx_1...dx_n \end{align*}

The intuition why I believe that this is true is the following:

$f^-$ may not be always smaller than $f^+$ but independently of how late we start the integral, the integral over $f^+$ remains bigger. So independently of how late we start, $f^+$ is "most of the times bigger than $f^-$ or it is drastically bigger than $f^-$ sometimes". So what is the worst that multiplying with $h$ can do? That is, that when for a while $f^-$ happens to be bigger than $f^+$ for these values $h$ should be huge. So if we hadn't the monotonicity of $h$ then we could make $h$ huge on the described sets and small otherwise. But here every increase in $h$ ultimately increases the value of the integral with $f^+$ because $h$ can't become smaller again, and in the long run, $f^+$ is bigger more often / drastically. Thus I think the claim should be true.