Well, assume your space is $([0,1],\mathcal{B},\lambda)$ with the latter denoting the Borel-algebra and the Lesbegue measure, and let $Y(x)=\frac{1}{x}$ and consider the convex function $f(x)=\frac{1}{x}$. Then, $f(Y(x))=x,$ which is clearly in $L^1$. However, $Y$ isn't.
Conversely, taking $f$ to be the same $Y(x)=x$ shows that the other way around might fail as well.
Thus, it is an important assumption that both $Y$ and $f(Y)$ are $L^1$-variables, since otherwise, either of the sides of the inequality might not even be defined.
I guess you are reading Measure Theory and Fine Properties of Functions - Evans, and your question is in page25.
Given the "i)" -- "Decomposition of nonnegative measurable functions" in page19 you listed above, it suffices to prove the question. The basic idea of the proof is to use the decomposition to prove the inequality:
$$\int_{*} fd\mu \geq \int^{*} fd\mu$$
but it's a little tricky, here is the proof:
1)$\mu(f=+\infty)>0\ $or$\ \exists n\in N_{+}\ s.t.\ \mu(A_n)=+\infty$
If $\mu(f=+\infty)>0$, any simple function $g$ s.t. $g\geq f, \mu-a.e.$ must have $\mu(g=+\infty) > 0$, and it implies:
$$\int^{*}fd\mu \geq \int_{*}fd\mu \geq (+\infty) * \mu(g=+\infty) = +\infty$$
$$\Rightarrow \int^{*}fd\mu = \int_{*}fd\mu = +\infty$$
therefore f is integrable.
If $\exists n\in N_{+}\ s.t.\ \mu(A_n)=+\infty$, simple function $\frac{1}{n}\chi_{A_n} \leq f$, and therefore:
$$\int^{*}fd\mu \geq \int_{*} fd\mu \geq \int \frac{1}{n}\chi_{A_n} d\mu = \frac{1}{n} * \mu(A_n) = +\infty$$
$$\Rightarrow \int^{*}fd\mu = \int_{*} fd\mu = +\infty$$
which implies $f$ is integrable.
2)$\mu(f=+\infty)=0\ $and$\ \forall n\in N_{+}\ \mu(A_n)<+\infty$
Now we can dismiss the part that $f=+\infty$ and only consider the bounded part of $f$, since the simple funcitons required in the lower and upper integral only require a.e. inequality.
Given the decomposition:
$$f=\sum_{n=1}^{+\infty} \frac{1}{n} \chi_{A_n}$$
we have:
$$\sum_{n=1}^k \frac{1}{n} \chi_{A_n} \leq f \quad \forall k \in N_{+}$$
and $\sum_{n=1}^k \frac{1}{n} \chi_{A_n}$ is simple function by definition, so we have:
$$\begin{align}
\int_{*}fd\mu &\geq \int \sum_{n=1}^k \frac{1}{n} \chi_{A_n} d\mu \\\\
& = \sum_{n=1}^{g(k)} y_n\mu(f^{-1}(B_n)) \quad \forall k \in N_{+}
\end{align}$$
where $B_n = \{x|f(x)=y_n, y_n\in Image(\sum_{n=1}^k \frac{1}{n} \chi_{A_n})\}, g(k)\subseteq N_{+}, g(k)\geq k\ $ ($\{A_n\}$ are not disjoint), and $\mu(B_n)<+\infty$, since $B_n \subseteq \cup_{n=1}^k A_n$ and $\forall n\leq k,\ \mu(A_n)<+\infty$
let $k \rightarrow +\infty$, we have:
$$\int_{*}fd\mu \geq \sum_{0\leq y_n\leq +\infty} y_n\mu(f^{-1}(B_n))$$
where $B_n = \{x|f(x)=y_n, y_n\in Image(\sum_{n=1}^{+\infty} \frac{1}{n} \chi_{A_n})\}$
Since $f < +\infty \ a.e.$ and $f=\sum_{n=1}^{+\infty} \frac{1}{n} \chi_{A_n}$, we have:
$$\forall \epsilon > 0\ \exists N\in N_{+}\ s.t.\ \forall k>N\quad f\leq \epsilon + \sum_{n=1}^k \frac{1}{n}\chi_{A_n}\quad a.e.\ \forall k \in N_{+}$$
therefore:
$$\begin{align} \int^{*}fd\mu &\leq \int \epsilon + \sum_{n=1}^k \frac{1}{n}\chi_{A_n}d\mu \\\\ &= \sum_{n=1}^{g(k)} y_n\mu(f^{-1}(B_n)) + \sum_{n=1}^{g(k)} \epsilon \mu(f^{-1}(B_n)) \quad \forall k \in N_{+} \end{align}$$
let $\epsilon \rightarrow 0^{+}$, we have:
$$\int^{*}fd\mu \leq \sum_{n=1}^{g(k)} y_n\mu(f^{-1}(B_n))\quad \forall k \in N_{+}$$
then let $k\rightarrow +\infty$, we have:
$$\int^{*}fd\mu \leq \sum_{0\leq y_n\leq +\infty} y_n\mu(f^{-1}(B_n))$$
Combine two inequalities above, we have:
$$\int_{*} fd\mu \geq \sum_{0\leq y_n\leq +\infty} y_n\mu(f^{-1}(B_n))\geq \int^{*}fd\mu$$
which implies $f$ is integrable.
I hope you find this proof helpful.
Best Answer
$\varphi \circ g$ always has integrable negative part so that the integral on the right hand side is well-defined via $$\int \varphi \circ g d \mu = \int (\varphi \circ g)^+ d\mu - \int(\varphi \circ g)^- d\mu$$
To prove this, note that convex functions possess subderivatives so that there are numbers $a,b$ such that $$ax + b \leq \varphi(x)$$
In particular, $(\varphi \circ g) \geq a g + b$ so that $$0 \leq (\varphi \circ g)^{-} \leq |a g + b| \in L^1$$