Exercise 1.10

Question

Suppose $\mu(X) < \infty$, $\{f_n\}$ is a sequence of bounded complex
    measurable functions on $X$, and $f_n 	o f$ uniformly on $X$. Prove that
    \begin{equation*}
      \lim_{n 	o \infty} \int_{X} f_n\,d\mu = \int_{X} f\,d\mu,
    \end{equation*}
    and show that the hypothesis ``$\mu(X) < \infty$'' cannot be omitted.

Amit Awasthi · Accepted Answer

\begin{proof} Let $\epsilon > 0$ be given, and let $N$ such that $\lvert f_n(x) - f(x) vert < \epsilon/\mu(X)$ for all $n \ge N$ and for all $x \in X$. Then, we have \begin{equation*} \left\lvert \int_X (f_n - f)\,d\mu ight vert \le \int_X \lvert f_n - f vert\,d\mu < \epsilon. \end{equation*} \par Now let $f_n = \chi_{[0,n]}/n$. Then, $f_n o 0$ uniformly, since given $\epsilon > 0$, for all $n > 1/\epsilon$, we have $\lvert f_n(x) vert \le 1/n < \epsilon$. On the other hand, \begin{equation*} \lim_{n o \infty} \int_X f_n\,d\mu = 1 e 0 = \int_X 0\,d\mu.\qedhere \end{equation*} \end{proof}

Exercise 1.10

Answers

Comments

Exercise 1.10

Answers

Comments

Add answer