 Weak law without an absolute first moment

Let $X$ have a symmetric probability distribution with $\mathbb{P}(|X|>x)=1/(x\log x)$ for $x\ge e$, mass $1-1/e$ at zero, and no mass with $0<|X|<e$. Then $\mathbb{E}|X|=\infty$, but independent copies satisfy the weak law of large numbers with limit zero. Indeed $n\mathbb{P}(|X|>n)=1/\log n\to 0$. Their symmetric truncations $Y_{n,j}=X_j{1}_{\{|X_j|\le n\}}$ have mean zero and second moment $O(n/\log n)$, so Chebyshev's inequality gives $n^{-1}{\sum }_{j\le n}{Y}_{n,j}\to 0$ in probability. A union bound shows that the original and truncated sums differ with probability tending to zero. This also gives a differentiable characteristic function at zero whose derivative cannot be interpreted as an absolutely convergent expectation.