Exercise 11.6

Question

Exercise 6: Let
    $$
        f_n(x)=
        \begin{cases}
            \displaystyle\frac{1}{n} &(|x|\le n) \
            0 & (|x|>n).
        \end{cases}
    $$
    Then $f_n(x)\rightarrow 0$ uniformly on $\mathbf R$, but
    $$
        \int_{-\infty}^\infty f_n\,dx=2\quad n=1,2,\ldots.
    $$
    Thus uniform convergence does not imply dominated convergence in the sense of Theorem 11.32. However, on sets of finite measure, uniformly convergent sequences of bounded functions do satisfy
    Theorem 11.32.

Amit Awasthi · Accepted Answer

\begin{proof}
      The integral being $2$ is obvious. For uniform convergence, note that
      given any $\epsilon > 0$, letting $N$ be such that $1/N < \epsilon$, we
      have that $\lvert f_n(x) - f_m(x) \rvert < 1/N$ for any $m,n \ge N$ and
      $x \in E$.
    \end{proof}

Exercise 11.6

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Exercise 11.6

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