Exercise 7.4.7

Question

Review the discussion immediately preceding Theorem 7.4.4.
\begin{enumerate}[label=(\alph*)] 
\item Produce an example of a sequence $f_{n} \rightarrow 0$ pointwise on $[0,1]$ where $\lim _{n \rightarrow \infty} \int_{0}^{1} f_{n}$ does not exist.
\item Produce an example of a sequence $g_{n}$ with $\int_{0}^{1} g_{n} \rightarrow 0$ but $g_{n}(x)$ does not converge to zero for any $x \in[0,1]$. To make it more interesting, let's insist that $g_{n}(x) \geq 0$ for all $x$ and $n$.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item $$f_n(x) = \begin{cases}
   n^2 & x \in (0, 1/n) \
   0 & \text{otherwise}
\end{cases}$$
\item Define the set $S_{i,j} = [(i-1)/j, i/j]$ for $j > i > 0$, and the sequence of sets $R_i$ enumerating through all $S_{i,j}$; specifically $R_1 = S_{1,1},\ R_2 = S_{1,2},\ R_3 = S_{2,2},\ R_4 = S_{1,3}$, and so on. Let
$$g_n(x) = \begin{cases}
    1 & x \in R_n \
    0 & \text{otherwise}
\end{cases}$$
Each $g_n$ looks like a ``pulse'', and as $n$ increases and we increase the $j$ index of $S_{i,j}$, the pulse gets increasingly narrower; in particular, if $g_n$ is based on $S_{i,j}$ then $\int^1_0 g_n = 1/j$. This ensures $\lim_{n \to \infty} \int^1_0 g_n = 0$. However, as we go through the $i$ index, we slide the pulse over all values in $[0,1]$. A bit more formally, for all $x \in [0,1]$ and for all $j > 0$, $x \in S_{i,j}$ for some $i$. This ensures that for all $x \in [0,1]$, $g_n(x) = 1$  infinitely many times, preventing $g_n(x) \to 0$.
 \end{enumerate}

Exercise 7.4.7

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Exercise 7.4.7

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