Exercise 11.3

Question

\def\ints{\mathbb{Z}}
Exercise 3: If $\{f_n\}$ is a sequence of measurable functions, prove that the set of points $x$ at which $\{f_n(x)\}$ converges is measurable.

analambanomenos · Accepted Answer

\def\where{\,|\,}
\def\ints{\mathbb{Z}}

If $h$ is a measurable function, then for every real number $a$
    $$
        h^{-1}(a)=\{x\,|\,h(x)=a\}=\{x\,|\,h(x)\ge a\}\cap\{x\,|\,h(x)\le a\}
    $$
    is a measurable set by Theorem 11.15. Also, if $g_1$ and $g_2$ are measurable functions then $h=g_1-g_2$ is also a measurable function by Theorem 11.18. Hence $h^{-1}(0)$, the set where
    $g_1(x)=g_2(x)$, is a measurable set.

Letting
    $$
        g_1(x)=\limsup_nf_n(x)\quad\quad g_2(x)=\liminf_nf_n(x),
    $$
    then these are measurable functions by Theorem 11.17, hence the set where $g_1(x)=g_2(x)$ is measurable. But this is precisely the set of points at which $\{f_n(x)\}$ converges.

Amit Awasthi · Answer

\begin{proof}
      We can write this set as
      \begin{equation*}
        \bigcap_{k=1}^\infty \bigcup_{N=1}^\infty \bigcap_{m,n \ge N}
        \{x \in E \mid \lvert f_m(x) - f_n(x) \rvert < 1/k\}.\qedhere
      \end{equation*}
\end{proof}

Exercise 11.3

Answers

Comments

Comments

Exercise 11.3

Answers

Comments

Comments

Add answer