We are given that $g\in L^1([0,1])$ amd $\|g\|_1 > 0$. Additionally, we have a function $f\in L^{\infty}([0,1])$ that is nonnegative, has $\|f\|_{\infty} > 0$, and has $\mu(E) = 0$ where $E = \{x:f(x) = \|f\|_{\infty}\}$. I need to show that
$$ \int fg < \|g\|_1\|f\|_{\infty}.$$
Here's what I have so far. We let $A = \{x:f(x) > \|f\|_{\infty}\}$ and $B = \{x: f(x) < \|f\|_{\infty}\}$, then $[0,1] = A \cup B \cup E$, and $\mu(A) = \mu(E) = 0$. So
$$ \int fg = \int_{E^c} fg = \int_B fg.$$
Then we know that on $B$, $f < \|f\|_{\infty}$, so I want to say that
$$ \int_B fg < \int_B\|f\|_{\infty}g,$$
but I don't know that this is true. If this were true, then I would essentially be done, but I'm not sure whether this actually holds. Any suggestions? I haven't used the fats that $g\in L^1$ and $f\in L^{\infty}$, nor the facts that $\|g\|_1 < \infty$ and $\|f\|_{\infty} < \infty$. Perhaps these are useful?