Product of a non negative decreasing function with a Lebesgue integrable function on $[0, 1]$

Question

Suppose $f$ is a Lebesgue integrable function on $[0, 1]$ such that $$\int^t_0f(x)dx \leq 0$$ for all $t \in [0, 1]$ and $\phi : [0, 1] \rightarrow \mathbb{R}$ is a non-negative decreasing function then show that $\phi f$ is Lebesgue integrable on $[0, 1]$ and $$\int^t_0f(x)\phi(x)dx \leq 0 - (*)$$ for all $t \in [0, 1]$. $$$$ I showed that $f\phi$ is Lebesgue integrable on $[0, 1]$ but I'm unable to prove $(*)$.

Answer 1

This is just an tweak of Did's answer.

Define a sequence of step functions by letting $\displaystyle I_k=\left [{k\over n},{k+1\over n}\right )$ and $\displaystyle \phi _n=\sum \limits _k\phi \left ({k\over n}\right ){\bf 1}_{I_k}$. We have $0\leq \phi _n(x)\leq \phi (0)$ for all $x,n$, and $\phi _n(x)\downarrow \phi (x)$ for ae. $x\in [0,1]$.

The Dominated Convergence Theorem shows that $\displaystyle \int f\phi _n\to \int f\phi$.

We can write $\displaystyle \phi _n(x)=\mu [x,1]=\int \limits _x^1d\mu$ for some measure $\mu$. To see this, let $\displaystyle \mu _n=\phi (1)\delta _1+\sum _{k=0}^{n-1}\left (\phi \left ({k\over n}\right )-\phi \left ({k+1\over n}\right )\right )\delta _{k \over n}$, where $\delta _y$ is the Dirac measure at $y$. Then \begin{eqnarray} \int \limits _0^tf(x)\phi _n(x)\,dx & = & \int \limits _0^tf(x)\int _x^1d\mu (s)\,dx \\ & = & \int \limits _0^1\int _0^1{\bf 1}_{[0,t]}(x){\bf 1}_{[x,1]}(s)f(x)\,d\mu (s)\,dx \\ & = & \int \limits _0^1\int \limits _0^1{\bf 1}_{[0,1]}(s){\bf 1}_{[0,\min(s,t)]}(x)f(x)\,dx\,d\mu(s) \\ & = & \int \limits _0^1\left (\int \limits _0^1{\bf 1}_{[0,1]}(s){\bf 1}_{[0,\min(s,t)]}(x)f(x)\,dx\right )\,d\mu (s) \\ & = & \int \limits _0^1\left (\int \limits _0^{\min (s,t)}f(x)\,dx\right )\,d\mu (s) \\ & \leq & 0. \end{eqnarray} Then the result follows from the Dominated Convergence Theorem.