Exercise 2.9

Question

Construct a sequence of continuous functions $f_n$ on $[0,1]$ such that $0
    \le f_n \le 1$, such that
    \begin{equation*}
      \lim_{n 	o \infty} \int_0^1 f_n(x)\,dx = 0,
    \end{equation*}
    but such that the sequence $\{f_n(x)\}$ converges for no $x \in [0,1]$.

Amit Awasthi · Accepted Answer

\begin{proof}
      Let $E_n = \lfloor [\sum_1^n 1/k, \sum_1^{n+1} 1/k] floor$, and let
      $f_n$ be a modification of $\chi_{E_n}$ such that in the graph of $f_n$,
      the images of the two endpoints $\lfloor \sum_1^n 1/k floor$ and
      $\lfloor \sum_1^{n+1} 1/k floor$ of $E_n$ are connected to the $x$-axis
      by line segments of slope $n$. Then, we have
      \begin{equation*}
        \int_0^1 f_n(x)\,dx = \frac{2}{n} 	o 0\ 	ext{as}\ n 	o \infty.
      \end{equation*}
      However, $\{f_n(x)\}$ does not converge for any $x \in [0,1]$ since every
      $x \in [0,1]$ is contained in infinitely many set of the form $E_n$,
      but is also contained in infinitely many sets of the form $E_n^c$.
\end{proof}

Exercise 2.9

Answers

Comments

Exercise 2.9

Answers

Comments

Add answer