For $f \in L^1(a,b)$ show there is a unique $g$ such that $\int_a^x g(t)\, dt = (\int_a^x f(t) \, dt)^2$ I encountered this problem in self-study (reviewing of absolute continuity) of Jones, Lebesgue Integration on Euclidean Space, Ch 16 #30:
Assume $f \in L^1(a,b)$. Show that there exists a unique $g \in L^1(a,b)$ such that $$\int_a^x g(t) \, dt = \left( \int_a^x f(t) \, dt \right)^2$$
I have posted my own solution and would appreciated any comments about it's correctness, as well as any better ways to go about it.

Put $g(x) = 2 f(x) \int_a^x f(t) dt$. $g \in L^1(a,b)$ since $f \in L^1(a,b)$ and $F(x) = \int_a^x f(t) dt$ is absolutely continuous, hence continuous and bounded on $[a,b]$.
Then we can rewrite the above as 
\begin{align*}
g(x) &= f(x) \int_a^x f(t)dt + \int_a^x f(t)dt \cdot f(x) \\
\frac{d}{dx}\int_a^x g(t)dt &= \frac{d}{dx} \left[ \left( \int_a^x f(t)dt \right)^2 \right] \\
\implies \int_a^x g(t)dt &= \left[ \left( \int_a^x f(t)dt \right)^2 \right]
\end{align*}
If $\bar{g}$ purports to be another such function, then the same calculation (in reverse) will show it it identical to $g$, hence $g$ is unique.
