For every continuous unbounded function $g$ on $(0,1)$ there exists an integrable nonnegative function $f$ such that $g \circ f$ is not integrable $(0,1)$.
I've been working on the above problem for a few days now without much progress. If $g(x)=1/x$, then $f(x) = x$ works. If $g(x) = 1/\sqrt{x}$, then take $f(x) = x^2$, etc….My thought was to show that there exists an $n \in \Bbb{N}$ such $g \circ f$ is not integrable with $f(x) = x^n$.
I'm not sure if this is true, however…I could use a hint.
Best Answer
Suppose $x_n$ is a sequence in $(0,1)$ where $|g(x_n)| > 2^n$. Define a function $f$ on $(0,1)$ so that $f(x) = x_n$ on $(2^{-n-1}, 2^{-n}]$.