If $f:(0,\infty)\rightarrow\ (0,\infty)$ is a measurable function such that $\int_{0}^{\infty}f(x)\ dx=c$... 
If $f:(0,\infty)\rightarrow\ (0,\infty)$ is a measurable function such that, $$\int_{0}^{\infty}f(x)\ dx=c$$
Prove that, $$\lim_{n\rightarrow\infty}\int_{0}^{\infty}n \log(1+(\frac{f(x)}{n})^a)=\ \begin{cases} 
      0,&if\  a>1\\
      c,&if\  a=1 \\
      \infty,&if\  a<1 
   \end{cases}
\
$$

$\underline{a=1}$
Consider
$\lim_{n\rightarrow\infty}\int_{0}^{\infty}\textbf[n \log(1+(\frac{f(x)}{n})\textbf]-f(x)$
$f(x)=\log(\exp(f(x))=\log(\lim_n(1+\frac{f(x)}{n})^n)$
Then
$\lim_{n\rightarrow\infty}\int_{0}^{\infty}\textbf[n \log(1+(\frac{f(x)}{n})\textbf]-\log(\lim_n(1+\frac{f(x)}{n})^n)=\lim_n\int\lim_n \log(1+\frac{f(x)}{n})-\log(1+\frac{f(x)}{n})=0$
we can bring the limit outside because $n \log(1+(\frac{f(x)}{n})$ converges after L'Hopital
can you confirm this ?
 A: Suppose $\alpha < 1$. Let $L_B = \{ x | f(x) \le B \}$. For some $B$, we have $\int_{L_B} f \ge {c \over 2} >0$.
Consider $\phi(t) = t^\alpha$. For all $K>0$ there is some $\delta_K >0$ such that $\phi (t) \ge Kt$ for all $t\in [0,\delta_K]$.
Choose $K>0$, and pick $n$ large enough so that if $x \in L_B$ then ${f(x) \over n} \le \delta_K$.
Then
\begin{eqnarray}
\int n \log(1+({f(x) \over n})^\alpha) dx \
&\ge& \int_{L_B} n \log(1+({f(x) \over n})^\alpha) dx \\
&\ge& \int_{L_B} n \log(1+K{f(x) \over n}) dx \\
&\ge& \int_{L_B}  \log(1+K{f(x) \over n})^n dx
\end{eqnarray}
Fatou's lemma gives
\begin{eqnarray}
I &=& \liminf_n\int n \log(1+({f(x) \over n})^\alpha) dx \\
&\ge& \int \liminf_n ( n \log(1+({f(x) \over n})^\alpha) )dx \\
&\ge& \int_{L_B} \liminf_n ( \log(1+K{f(x) \over n})^n )dx \\
&=& \int_{L_B}K f(x) dx \\
&\ge& K {c \over 2}
\end{eqnarray}
Since $K$ was arbitrary, we have $I = \infty$.
Now suppose $\alpha \ge 1$. Note that for $x\in (0,1)$ we have
$1+ x^\alpha \le 1 + x \le (1+x)^\alpha$. For $x\ge 1$, we have
$\frac{1+x^\alpha}{(1+x)^\alpha} = \frac{1+ \frac{1}{x^\alpha} }{ (1+ \frac{1}{x})^\alpha } \le \frac{1+ \frac{1}{x^\alpha} }{ 1+ \frac{1}{x}} \le 1 $,
hence we have $1+ x^\alpha \le (1+x)^\alpha$ for all $x >0$.
This gives $n \log(1+({f(x) \over n})^\alpha ) = n \log ( 1+{f(x) \over n})^\alpha = n \alpha \log ( 1+{f(x) \over n}) \le \alpha f(x)$. Since $f$ is integrable, we can apply the dominated convergence theorem.
If $\alpha = 1$, then $(1+{f(x) \over n} )^n \to e^{f(x)}$, and so we get 
$\int n \log(1+{f(x) \over n}) dx \to \int f(x) dx = c$.
If $\alpha >1$ and $t \ge 0$, then
$(1+({t \over n})^\alpha)^n = $
$(1+{t^\alpha \over n^\alpha} )^n = \left( (1+{t^\alpha \over n^\alpha} )^{n^\alpha} \right)^{1 \over n^{\alpha-1}} \le (e^{t^\alpha})^{1 \over n^{\alpha-1}} = e^{t^\alpha \over n^{\alpha-1}}$. Hence $ (1+({f(x) \over n})^\alpha)^n \to 1$, from which we get
$ \int n \log(1+({f(x) \over n})^\alpha) dx \to 0$.
