A random variable need not have a finite expectation

Define $\mathbb{P}(A):={\sum }_{n\in A}{p}_n(A\subseteq \mathrm{\Omega })\mathrm{.}$ The total mass is ${\sum }_{n=1}^{\mathrm{\infty }}\frac{1}{n(n+1)}={\sum }_{n=1}^{\mathrm{\infty }}(\frac{1}{n}-\frac{1}{n+1})=1.$ If $(A_j{)}_{j\ge 0}$ is pairwise disjoint, regrouping the nonnegative series gives $\mathbb{P}\text{\hspace{0.17em}⁣}\left({\bigcup }_{j\ge 0}{A}_j\right)={\sum }_{j\ge 0}{\sum }_{n\in A_j}{p}_n={\sum }_{j\ge 0}\mathbb{P}(A_j)\mathrm{.}$ Thus $\mathbb{P}$ is a probability measure on the power set of $\mathrm{\Omega }$, and $X(n)=n$ is a measurable real random variable.

Put $X_N:=X{\mathbf{1}}_{\{1,\dots ,N\}}$. The functions $X_N$ increase pointwise to $X$. By [L1], [L2], and monotone convergence, $\mathbb{E}[X]={\lim }_{N\to \mathrm{\infty }}\mathbb{E}[X_N]={\lim }_{N\to \mathrm{\infty }}{\sum }_{n=1}^{N}n\frac{1}{n(n+1)}={\sum }_{n=1}^{\mathrm{\infty }}\frac{1}{n+1}=+\mathrm{\infty }\mathrm{.}$ Thus the expectation exists only as an extended value, not as a finite real number, exactly as [L1] allows.

Steps 1.1 and 2.1 refute the claim that every random variable has finite expectation.