About the essential supremum of a function Let $f$ be an everywhere finite measurable function of a measure space $(\Omega, \Sigma, \mu)$, such that for every continuous function $\alpha: \mathbb R \to \mathbb R$, the composition $\alpha\circ f$ is integrable. Prove that $||f||_{\infty} < \infty$.
Here are my ideas so far: Assume without loss of generality that $f \ge 0$. Let $g_n = f^n/n!$. Each $g_n$ is integrable. By the dominated convergence theorem, 
$$
\Sigma_{n=1}^\infty \int f^n/n!\, \mathrm d\mu= \int e^f\,\mathrm d\mu  
$$
The right-hand-side is finite. So $\lim_{n\rightarrow\infty} \int f^n/n!\, \mathrm d\mu = 0$. 
Now, suppose for a contradiction that $\mu(\{f>n\}) >0$ for all $n \in N$. Then, by Cheybychev's inequality, we have 
$$
\mu(\{f>n\}) \le \frac{1}{n^n}\int f^n\,\,\mathrm d\mu
$$
Therefore, we have
$$
\int \frac{f^n}{n!}\,\mathrm d\mu \ge \frac{n^n \mu(\{f>n\})}{n!} 
$$
From here, I would like to derive a contradiction, since the right-hand-side of the above inequality seems to go infinity, contradicting that $\lim_{n\rightarrow\infty} \int f^n/n!  \,\mathrm d\mu = 0$. But I am not really confident my last claim, as I don't  understand the behavior of $\dfrac{n^n \mu(\{f>n\})}{n!}$.
A possible useful fact I found along the way is that $g_0$ is integrable implies that the space $\Omega$ has finite measure. 
 A: Let $\alpha \equiv 1.$ Then $\alpha\circ f\equiv 1$ is integrable. This implies $\mu(X)<\infty.$
Suppose, to reach a contradiction, that $f\notin L^\infty.$ For $n=1,2,\dots$ set $E_n = \{x: |f(x)|\in [n,n+1]\}$ Then $\mu(E_n)>0$ for infinitely many $n,$ say for $n_1 < n_2 < \cdots.$ By thinning out the subsequence, we can assume $n_{k+1}-n_k\ge 2$ for all $k.$
For $k=1,2,\dots$ define
$$\alpha (x)  = \frac{1}{\mu(E_{n_k})},\, |x|\in [n_k,n_k+1].$$
Then extend $\alpha$ any way you like to be continuous and nonnegative on $\mathbb R.$ Then
$$\int_X \alpha\circ f \, d\,\mu \ge \int_{\bigcup E_{n_k}} \alpha\circ f \, d\,\mu = \sum_{k=1}^{\infty}\int_{E_{n_k}}\alpha \circ f \, d\,\mu$$ $$ = \sum_{k=1}^{\infty}\frac{1}{\mu(E_{n_k})}\cdot \mu(E_{n_k}) = 1 + 1 + 1 +\cdots =\infty.$$
This is a contradiction. Hence $f \in L^\infty$ as desired.
This is a bit brief, so ask questions if you like.
A: Hint: consider an increasing sequence $N_n$ with $N_n > n$ and $\mu(\{N_n < |f| \le N_{n+1}\}) > 0$.  Construct a continuous function $\alpha$ that grows so rapidly 
that ....  But don't try for a neat formula.
A: Without further assumptions on $f$ this seems not to be correct. Choose some countable set $A\subset\Omega$, some enumeration $a_1,a_2,\ldots,a_n,\ldots$ of $A$ and set $f$ by $f(x)=0$ for all $x\not\in A$ and $f(a_n) = n$ else. Then $f$ is not zero on a countable set of measure zero, and clearly all the integrals mentioned exist. But $f$ is unbounded.
