Exercise 1.7

Question

Suppose $f_n \colon X 	o [0,\infty]$ is measurable for $n = 1,2,3,\ldots$,
    $f_1 \ge f_2 \ge f_3 \ge \cdots \ge 0$, $f_n(x) 	o f(x)$ as $n 	o \infty$,
    for every $x \in X$, and $f_1 \in L^1(\mu)$. Prove that then
    \begin{equation*}
      \lim_{n 	o \infty} \int_X f_n\,d\mu = \int_X f\,d\mu
    \end{equation*}
    and show that this conclusion does \emph{not} follow if the condition ``$f_1
    \in L^1(\mu)$'' is omitted.

Amit Awasthi · Accepted Answer

\begin{proof}
      Let $g_n = f_1 - f_n \ge 0$. Then, $g_1 \le g_2 \le \cdots$ hence by the
      monotone convergence theorem,
      \begin{equation*}
        \lim_{n 	o \infty} \int_X (f_1 - f_n)\,d\mu = \int_X (f_1 - f)\,d\mu
      \end{equation*}
      and subtracting $\int_X f_1\,d\mu$ from both sides we are done.
      \par On the other hand, suppose $X = \mathbf{R}$ and $f_n =
      \chi_{[n,\infty)}$. Then,
      \begin{equation*}
        \lim_{n 	o \infty} \int_X f_n\,d\mu = \infty 
e 0 = \int_X
        f\,d\mu.
      \end{equation*}
    \end{proof}

Exercise 1.7

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Exercise 1.7

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