Question: Let $E$ be a measurable subset of $\mathbb{R}$ and $f,g \in L^3(E)$ satisfying the following equality $$\|f\|_3 = \|g\|_3 = \int_E f^2 g =1.$$ Prove that $$g(x) = |f(x)|$$ for almost all $x \in E.$
My attempt:
By Holder's inequality, the integral $\int_E f^2g$ is defined.
I would like to show that the set $\{x\in E: g(x)=|f(x)|\}$ has measure zero.
By assumption, we have $$\int_E |f|^3 = 1, \int_E |g|^3 = 1 \text{ and } \int_E f^2g = 1.$$ Denoting $$\{x\in E: g(x) = |f(x)|\} = \{g = |f|\},$$ by additivity of integration, we have $$1 = \int_E |f|^3 = \int_{\{g=|f|\}}|f|^3 + \int_{\{g\neq |f|\}}|f|^3,$$ $$1 = \int_E |g|^3 = \int_{\{g=|f|\}}|f|^3 + \int_{\{g\neq |f|\}}|g|^3,$$ and $$1 = \int_E f^2 g = \int_{\{g=|f|\}}|f|^3 + \int_{\{g\neq |f|\}}f^2 g.$$ It follows that $$\int_{\{g\neq |f|\}}|f|^3 = \int_{\{g\neq |f|\}}|g|^3,$$ $$ \int_{\{g\neq |f|\}}|g|^3= \int_{\{g\neq |f|\}}f^2 g$$ and $$\int_{\{g\neq |f|\}}|f|^3 = \int_{\{g\neq |f|\}}f^2 g.$$ However, I could not proceed from here. Any hint would be appreciated.
Hints: Recall Young's inequality which says that if $p^{-1}+q^{-1}=1$, then $ab\leq \frac{a^p}{p}+\frac{b^q}{q}$, for all $a,b\geq 0$. Moreover, equality holds iff $a^p=b^q$.
Applying this with $a=|f|^2, b=|g|, p=3/2,q=3$ and then integrating both sides, we find that $$\int |f^2g| \leq \frac{2}{3}\int|f|^3+\frac{1}{3}\int|g|^3=1=\int f^2g \leq \int |f^2g|,$$ so all of the inequalities are equalities. Deduce that $f^2g= |f^2g| = \frac{2}{3}|f|^3+\frac{1}{3}|g|^3$ a.e. Now apply the condition for equality to hold in Young's inequality, then deduce that $g$ must be nonnegative on $E$ a.e.