Let $h:[0,\infty) \to \Bbb R$ measurable and $\int_0^\infty \vert h(y) \vert \text d y <\infty$ and suppose $$\int_0^c h(y) \text d y \geq 0 \quad \forall c >0$$ Let $\omega : [0,\infty ) \to [0,1] $ be a non-increasing function.
I want to show that under these assumptions (or maybe I need slightly more?) holds that $$\int_0^c h(y) \omega (y) \text d y \geq 0 \quad \forall c >0$$
EDIT: I added a $L^1$ condition, since otherwise it is not even clear if the integral is well-defined.
Since $\omega$ is non-increasing it is measurable and for $c\geq 0$ we can approximate $\omega$ on $[0,c]$ from below by simple functions $\omega_n$ of the form $\omega_n = \sum_{i=1}^n a_i \Bbb 1_{[t_{i-1} , t_i)}$, where $0 = t_0 < \ldots < t_n = c$ and $a_i \geq a_{i+1}$. Fix $n$ and suppose that for all $c\geq0$ and all simple functions of the form above with $n$ coefficients $a_i$ and lattice points $t_i$ holds that $$\sum_{i=1}^n a_i \int_{t_{i-1}}^{t_i} h(y) \text d y \geq a_n \int_0^{t_n} h(y) \text dy$$ Then holds for a simple function of above form with $n+1$ coefficients $b_i$ and lattice points $s_i$ that $$\sum_{i=1}^{n+1} b_i \int_{t_{i-1}}^{t_i} h(y) \text d y = \sum_{i=1}^n b_i \int_{t_{i-1}}^{t_i} h(y) \text dy +b_{n+1}\int_{t_n}^{t_{n+1}} h(y)\text dy\\ \geq b_n \underbrace{\int_0^{t_n} h(y) \text d y}_{\geq 0} + b_{n+1}\int_{t_n}^{t_{n+1}} h(y)\text dy\\ \geq b_{n+1} \int_0^{t_{n+1}} h(y) \text d y$$ Since the assumption above holds for $n=1$ by the fact that $\int_0^c h (y) \text d y \geq 0$, we have that for all $\omega_n$ holds $$\int_0^c h(y) \omega (y) \text d y \geq a_n \int_0^c h(y) \text dy \geq 0$$ Note that $\vert h \omega_n \vert \leq \vert h \vert \omega \leq \vert h \vert$, and $\vert h \vert$ is integrable by assumption. Thus $$\int_0^c h(y) \omega (y ) = \lim_{n\to\infty} \int_0^c h(y) \omega_n (y) \text dy \geq 0$$