Construct a sequence $(f_n)_n$ convergent in $L^{p}(\mathbb{R}^d)$ such that $\lim\limits_{n\rightarrow\infty}f_n(x)$ does not exist for a.e. $x\in\mathbb{R}^d$.
My thought so far:
I am trying to do this for the unit interval $[0,1]$ and then extending it to all of $\mathbb{R}$, and then $\mathbb{R}^d$. Take the following sequence of subintervals of $[0,1]$ $$[0,1],[0,1/2],[1/2,1],[0,1/3],[1/3,2/3],[2/3,1],[0,1/4],\dots,[3/4,1],\dots$$ Now denote the $n$th interval above by $E_n,$ and let $f_n(x)=1_{E_n}(x).$ Then $$\left(\int_{[0,1]}f_n(x)^{p}\,dx\right)^{1/p}=m(E_n)^{1/p}=\frac{1}{n^p},$$ and hence $f_n\rightarrow 0$ in the $L^p$-norm. However this sequence does not converge pointwise for any $x\in[0,1],$ since it oscillates between $0$ and $1.$
To extend this to $\mathbb{R}$, I take the following partition of the real line $$\bigcup_{z\in\mathbb{Z}}[z,z+1],\bigcup_{z\in\mathbb{Z}}[z,z+1/2],\bigcup_{z\in\mathbb{Z}}[z+1/2,z+1],\bigcup_{z\in\mathbb{Z}}[z,z+1/3],\bigcup_{z\in\mathbb{Z}}[z+1/3,z+2/3],\bigcup_{z\in\mathbb{Z}}[z+2/3,z+1],\dots,$$ which is a replication of the process done before for $[0,1].$ However, I think that in this construction we lose the convergence in the $L^p$-norm, since I believe that we now have, denoting the $n$th set above by $E_n$, $$\left(\int_{\mathbb{R}}1_{E_n}\,dx\right)^{1/p}=\left(\sum^{\infty}_{k=1}\frac{1}{n}\right)^{1/p},$$ which diverges to infinity
Is there any way to save the above argument? If not, can you provide a hint on how to construct such a sequence?
Thank you for your time, and appreciate your feedback.
Yes, it is possible to extend your idea to $\mathbb{R}$. Define intervals by $$ [-1,0], [0,1], \\ [-2,-\tfrac{3}{2}], [-\tfrac{3}{2},-1], [-1,-\tfrac{1}{2}],\ldots,[1,\tfrac{3}{2}],[\tfrac{3}{2},2],\\ [-3,-\tfrac{8}{3}],[-\tfrac{8}{3},-\tfrac{7}{3}],\ldots$$
If we denote the $n$-th interval by $I_n$, then the Lebesgue measure of $I_n$ tends to zero as $n \to \infty$, and hence $f_n := 1_{I_n}$ converges to $0$ in $L^p$ as $n \to \infty$. Because of the construction of the sequence, we know that for any $x \in \mathbb{R}$ we have $$\sharp \{n \in \mathbb{N}; f_n(x)=0\} = \infty \quad \text{and} \quad \sharp\{n \in \mathbb{N}; f_n(x)=1\} = \infty,$$ and therefore $f_n$ does not converge pointwise.
In $\mathbb{R}^d$ we can consider
$$g_n(x) := f_n(x_1) \exp \left( - \sum_{j=2}^d |x_j| \right), \qquad x=(x_1,\ldots,x_d) \in \mathbb{R}^d$$
where $f_n$ is the function from the first part of this answer.