Exercise 1.9

Question

Suppose $\mu$ is a positive measure on $X$, $f \colon X 	o [0,\infty]$ is
    measurable, $\int_X f\,d\mu = c$, where $0 <c < \infty$ and $\alpha$ is a
    constant. Prove that
    \begin{equation*}
      \lim_{n 	o \infty} \int_X n\log[1+(f/n)^\alpha]\,d\mu = \begin{dcases*}
        \infty & if 0 < \alpha < 1,\
        c & if \alpha = 1,\
        0 & if 1 < \alpha < \infty.
      \end{dcases*}
    \end{equation*}

Amit Awasthi · Accepted Answer

\begin{proof}
      Suppose $\alpha \ge 1$. Then, by the Taylor series for $e^x$, we have that
      $1 + x^\alpha \le e^{\alpha x}$. Substituting $x = f/n$ and applying
      $\log$, we get the inequality
      \begin{equation*}
        f_n \coloneqq n\log[1+(f/n)^\alpha] \le \alpha f.
      \end{equation*}
      Thus, we have that $\alpha f \ge f_n$ for all $n$. Now, by the dominated
      convergence theorem, we have
      \begin{equation*}
        \lim_{n 	o \infty} \int_X f_n\,d\mu = \int_X \lim_{n 	o \infty}
        f_n\,d\mu.
      \end{equation*}
      We now calculate $\lim_{n 	o \infty} f_n$:
      \begin{align*}
        \lim_{n 	o \infty} f_n &= \lim_{n 	o \infty} n\log[1+(f/n)^\alpha]\
        &= \lim_{n 	o \infty}
        \frac{\log[1+(f/n)^\alpha]^{n^\alpha}}{n^{\alpha-1}}.
      \end{align*}
      If $\alpha > 1$, then the numerator converges to $f^\alpha$ using the
      sequence definition of $e^x$, while the
      denominator diverges to $\infty$, hence the limit is zero. On the other
      hand, if $\alpha = 1$, then the denominator is $1$, and so the limit is
      $f^\alpha = f$. Thus, we have that the integral we are interested in has
      the desired values for $\alpha \ge 1$.
      \par Now suppose $\alpha < 1$. By Fatou's lemma, we have
      \begin{equation*}
        \int_X \liminf_{n 	o \infty} f_n\,d\mu \le \liminf_{n 	o \infty}\int_X
        f_n\,d\mu \le \lim_{n 	o \infty} \int_X f_n\,d\mu,
      \end{equation*}
      and it suffices to show the integral on the very left diverges. But we
      have
      \begin{align*}
        \liminf_{n 	o \infty} f_n &= \liminf_{n 	o \infty}
        n\log[1+(f/n)^\alpha]\
        &= \liminf_{n 	o \infty} n^{1-\alpha}\log[1+(f/n)^\alpha]^{n^\alpha}\
        &\ge f^\alpha\liminf_{n 	o \infty} n^{1-\alpha} = \infty
      \end{align*}
      everywhere $f(x) 
e 0$, which happens on a set of positive measure since
      we have $\int_X f\,d\mu > 0$.
    \end{proof}

Exercise 1.9

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Exercise 1.9

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