If $\mu$ is a finite measure on the measurable space $\big( X, \mathscr{F} \big)$, $f : X\to [ 0, +\infty)$ is measurable.
Then $\textbf{does it exist a continuous function $g : [ 0, +\infty)\to [ 0, +\infty) $ such that $g(x)$ increases to $+\infty$}$ as $x\to +\infty$ and $$\int_X g\big( f(x) \big) \, \mu (dx) < +\infty \,\, ? $$
W.l.g. we may assume $\mu(X) = 1$. Since $$\int_X g\big( f(x) \big) \, \mu (dx)= \int_0^\infty g(y) \, \nu(dy) $$ where $\nu(B) = \mu\circ f^{-1}(B)$ for $B$ Borel subset of $[0, +\infty)$, then it is equivalent to ask that given a nonnegative random variable $Y$, does it exist a continuous function $g : [ 0, +\infty)\to [ 0, +\infty) $ such that $g(x)$ increases to $+\infty$ as $x\to +\infty$ and $g(Y)$ is of finite expectation ?
If $\mu(X) = +\infty$, it is easy to come up with a counterexample :
Pick $f(x) = x$ defined on $\big( \mathbb{R}_+, \mathscr{B}(\mathbb{R}_+) , dx \big)$ and there is no continuous increasing function $g : \mathbb{R}_+ \to \mathbb{R}_+$ such that $\lim_{x\to+\infty} g(x) = +\infty$ and $g\circ f$ is integrable.
Since $\lim_{x\to+\infty} g(x) = +\infty$, for large $M > 0$, $g(x) > 1$ as $x\geq M$, then \begin{align*} \int_0^\infty g(x) \, dx = \int_0^M g(x) \, dx + \int_M^\infty g(x) \, dx \end{align*} note that the first integral is finite, while the second is infinite.
I'll do the second version. This is easy if $\nu$ has compact support, so suppose this is not the case.
Then we can fix a sequence $0=x_0 < x_1 < \ldots$, $x_n\to\infty$, so that with $I_n=[x_{n-1}, x_n)$, we have that $\mu(I_n)\le (1/2)\mu (I_{n-1})$: just choose the next $x_n$ so that the current interval contains at least $2/3$ of the remaining measure. So $\mu(I_n)\le 2^{-n}$, and we can now define an integrable $h$ by $h=2^{n/2}$ on $I_n$. Make this continuous and strictly increasing to obtain $g$, with $g\le h$.