= 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 $n\mathbb P(|X|>n)=1/\log n\to0$. Their symmetric truncations $Y_{n,j}=X_j\mathbf1_{\{|X_j|\le n\}}$ have mean zero and second moment $O(n/\log n)$, so <Chebyshev's inequality> gives $n^{-1}\sum_{j\le n}Y_{n,j}\to0$ 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.
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