2.12

First, define the sets: $$S_n^k = \Big[\frac{k-1}{2^n}, \frac{k}{2^n}\Big]$$ From these intervals, define the sequence of functions: $$f_n^k = \chi_{S_n^k}(x) + \chi_{S_n^k}(-x)$$ Where for each $n$ to progress $+1$, $k$ must sweep from $1$ to $2^n + 1$.

Certainly, for every $n$, $$\int_{\mathbb{R}} f_n^k = \frac{1}{2^{n-1}}$$ Since the collection of functions $\lbrace f_n^k \rbrace$ is countable, with a well-defined sequence, let us just enumerate them with the single index $m$. We have: $$ \int_{\mathbb{R}} |f_m - 0| = \int_{\mathbb{R}} f_m \to 0$$ However, for any $x \in \mathbb{R}$, it's clear that $\lim_{m\to \infty} f_m(x)$ does not exist. To expand this to $\mathbb{R}^d$, just replace the intervals with a closed balls centered at the origin subtracting open balls with the same dyadic rational-difference... (onion-layers).