multiplying the integrands in an inequality of integrals with same limits

Question

I am trying to find rigorous justification for the following.

If we have the following inequality $$ \int_{X}^{\infty}u(x)\mbox{d}x \le \int_{X}^{\infty}v(x)\mbox{d}x $$ If we now multiply both sides with $\exp(-k(x))$ as in $$ \int_{X}^{\infty}\exp(-k(x))u(x)\mbox{d}x \le \int_{X}^{\infty}\exp(-k(x))v(x)\mbox{d}x $$ then the inequality is preserved.

My initial thoughts were to collect the terms $$ 0 \le \int_{X}^{\infty}(v(x) - u(x))\mbox{d}x $$ and the then try use integration by parts after multiplying with $\exp(-k(x))$ to show the RHS will be satisfy the inequality, but that lead me to circular kind of argument. Also i do not know if $v(x)$ is differentiable.

How can this be rigorously justified ? If these conditions do not suffice, what additional conditions are minimally needed so this will hold.

Thank you

Answer 1

It is not true as it is stated at this moment. Take $u(x)=\mathbf{1}_{[0,1]}$ and $v(x)=\mathbf{1}_{[1,3]}$. It is clear that it satisfies the condition mentioned. Now take $k(x)=x$ and $X=0$ to get: \begin{align} I:=\int_0^\infty \exp(-x)u(x)\,dx=1-e^{-1} \end{align} while: \begin{align} J:=\int^\infty_0 \exp(-x)v(x)\,dx=e^{-1}-e^{-3} \end{align} Since $2e^{-1}<1$, we have $e^{-1}<1-e^{-1}$. Hence $e^{-1}-e^{-3}<1-e^{-1}$ and that implies $J<I$.

Answer 2

This is not true.

For an intuitive explanation of a counterexample: suppose $k(x)$ is zero "most of the time", so the inequality is unchanged, but then $k(x)$ happens to be very large for values of $x$ where $u(x)$ is zero, but it so happens that $v(x)$ is very large and negative.

Then $\int e^{-k(x)} v(x)$ will be much smaller, but $\int e^{-k(x)} u(x)$ would be unchanged, which can ruin the inequality.