Exercise 11.5

Question

Exercise 5: Put, for $0\le x\le 1$,
    \begin{align*}
        g(x) &=
              \begin{cases}
                  0 & \big(0\le x\le\frac{1}{2}\big) \
                  1 & \big(\frac{1}{2}<x\le 1\big)
              \end{cases} \
       f_{2k}(x) &= g(x) \
       f_{2k+1}(x) &= g(1-x)
    \end{align*}
    Show that for $0\le x\le 1$
    $$
        \liminf_{n\rightarrow\infty}f_n(x)=0
    $$
    but
    $$
        \int_0^1f_n(x)\,dx=\frac{1}{2}.
    $$

analambanomenos · Accepted Answer

For each $0\le x\le 1$, $\{f_n(x)\}$ is a sequence alternating between 0 and 1, so $\liminf f_n(x)=0$. Let $A=\big[0,\frac{1}{2}\big]$ and $B=\big(\frac{1}{2},1]$. Then, for $n$ even $f_n=K_B$,
    the characteristic function of $B$, and for $n$ odd $f_n=K_A$. Hence
    \begin{gather*}
        \int_0^1f_{2k}(x)\,dx=m(B)=\frac{1}{2} \
        \int_0^1f_{2k+1}(x)\,dx=m(A)=\frac{1}{2}
    \end{gather*}
    This shows that strict inequality is possible in the conclusion of Fatou's Lemma.

Amit Awasthi · Answer

\begin{proof}
      $\liminf_{n 	o \infty} f_n(x) = 0$ since for any $N$ and any $x \in E$,
      either $f_{2k}(x) = 0$ or $f_{2k+1}(x) = 0$ for all $2k+1 \ge N$. On the
      other hand, $\int_0^1 f_n(x)\,dx = \int_0^1 g(x)\,dx = \frac{1}{2}$ by
      construction.
    \end{proof}

Exercise 11.5

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Exercise 11.5

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