The norm $\|f_n-f\|_{L^1} \to 0$ but $f_n \not\to f$ A classmate and I are studying this following question from Stein-Shakarchi, Chapter 2, Exercise 12:

Show that there are $f \in L^1(\mathbb{R}^d)$ and a sequence $\{f_n\}$ with $f_n \in L^1(\mathbb{R}^d)$ such that $$\|f-f_n\|_{L^1(\mathbb{R}^d)} \to 0,$$ but $f_n(x) \to f(x)$ for no $x$.
[Hint: In $\mathbb{R}$, let $f_n=\chi_{I_n}$, where $I_n$ is an appropriately chosen sequence of intervals with $m(I_n)\to 0$.]

Our attempt:
First, I defined $I_n := [\frac{k-1}{2^n}, \frac{k}{2^n}]$ for all $k \in \mathbb{R}$, so that $m(I_n)\to 0$ as $n \to \infty$.
From these intervals, the sequence of functions is defined to be:
$$f_n := \chi_{I_n}(x) + \chi_{I_n}(-x)$$
Then $f=0$, and so
$$\int_{\mathbb{R}} |f_n - f| = \int_{\mathbb{R}} f_n = \frac{1}{2^{n}} +\frac{1}{2^{n}} \to 0.$$
But I am left to show that $f_n \to f$ for no $x$. I do not see this from my example. But does this example work? If so, why does $f_n$ not converge to $f$ for any $x$.
 A: In one dimension, consider in order 
$$\chi_{[0,1]}, \chi_{[0,1/2]}, \chi_{[1/2,1]},\chi_{[0,1/3]},\chi_{[1/3,2/3]},\chi_{[2/3,1]}, \dots $$
This sequence $\to 0$ in $L^1,$ and pointwise nowhere.
A: We will give an example in $\mathbb{R}^1$. The analagous construction in $\mathbb{R}^d$ is similar.
Fix $k \in \mathbb{Z}$. Take some enumeration $\{I^k_n\}_{n \in \mathbb{N}}$ of all subintervals of $[k,k+1]$ which have the form $[\frac{p}{q},\frac{p+1}{q}]$ for some integers $p,q$. Notice that $m(I^k_n) \to 0$ as $n \to \infty$, because for any $\epsilon >0$, there are only finitely many $n$ such that $m(I^k_n)>\epsilon$. Define $g^k_n=1_{I^k_n}$, for $n \in \mathbb{N}$. Notice that, for any $x \in [k,k+1]$ there are infinitely many $n$ with $g^k_n(x)=1$. Therefore $\limsup_{n \to \infty} g^k_n(x)=1$ for every $x \in [k,k+1]$.
For $n \in \mathbb{N}$, let $f_n = \sum_{k \in \mathbb{Z}} 2^{-|k|}g^k_n$, and let $f=0$. Then $\int |f_n -f|= \sum_{k \in \mathbb{Z}} 2^{-|k|}m(I^k_n)$, which converges to zero as $n \to \infty$ by the DCT (applied to counting measure), because each individual $m(I^k_n)$ approaches zero when $k$ is fixed and $n \to \infty$. Therefore $||f_n - f||_{L^1} \to 0$. However, if $x \in \mathbb{R}$, say $x \in [k,k+1)$ where $k \in \mathbb{Z}$, then $\limsup f_n(x)= 2^{-|k|}>0$, so $f_n(x)$ does not converge to $f(x)$, for any $x \in \mathbb{R}$.
