Let $f,g:[0,1]\to[0,M]$ be measurable for some $M<\infty$. We know by the Hoelder inequality that for any $p\geq1$, $$\|fg\|_1\leq\|f\|_p\|g\|_{1/(1-1/p)}\,.$$
Do we also have (for this special case of bounded functions on bounded domains) that $\|\cdot\|_1$ is actually submultiplicative, that is, $$\|fg\|_1\leq\|f\|_1\|g\|_1\,??$$
Here is a suggested proof for which I don't immediately see the problem: \begin{align*} \|fg\|_1 & \equiv \int_{[0,1]}fg \\ & = \int_{[0,1]^2}f(x)g(y)\delta(x-y)~\mathrm dx~\mathrm d y. \end{align*}
Now since as measures on $[0,1]^2$, $\delta(x-y)~\mathrm dx~\mathrm d y\leq\mathrm dx~\mathrm d y$, we find (using Fubini's theorem) the claim.
Take any set $E$ with $0<m(E)<1$ and take $f=g=I_E$. Then $\|fg\|_1=m(E)$ and $\|f\|_1\|g\|_1=m(E)^{2}$. But $m(E)^{2} <m(E)$.