Expectation does not always exist

Last updated Nov 1, 2022

Suppose $(\Omega, \mathcal{M}, \mathbb{P})$ is a Probability Space. There exists a Random Variable $X$ on $\Omega$ so that $X \not\in \bar{L^{1}}(\Omega)$. In other words, $\mathbb{E}(X)$ does not exist and we cannot even assign it a value of $\infty$ or $-\infty$ in the Compactification of the Real Numbers.

Consider the Probability Space $([-1, 1], \mathcal{B}([-1, 1]), \frac{1}{2}\lambda)$, where $\lambda$ is the Lebesgue Measure. Consider Random Variable

$$X = \begin{cases} \frac{1}{x} & \text{if } x \neq 0\\ 0 & \text{if } x=0\end{cases}$$ Then $$X^{+} = \begin{cases} \frac{1}{x} & \text{if } x > 0\\ 0 & \text{if } x \leq 0\end{cases}$$ and $$X^{-} = \begin{cases} -\frac{1}{x} & \text{if } x < 0\\ 0 & \text{if } x \geq 0\end{cases}$$ Then

$$\begin{align*} \frac{1}{2} \int\limits X^{+} d \lambda &= \frac{1}{2} \int\limits_{(0,1]} \frac{1}{x} d \lambda\\ &=\lim\limits_{\epsilon \downarrow 0}\left[-\frac{1}{2} \frac{1}{x^{2}}\right]{\epsilon}^{1} \\ &=-\frac{1}{2} + \lim\limits{\epsilon \downarrow 0}\frac{1}{2\epsilon^{2}}\\ &= \infty \end{align*} $$ Likewise $$\begin{align*} \frac{1}{2} \int\limits X^{-} d \lambda &= \frac{1}{2} \int\limits_{[-1,0)} -\frac{1}{x} d \lambda\\ &=\lim\limits_{\epsilon \downarrow 0}\left[\frac{1}{2} \frac{1}{x^{2}}\right]{-1}^{-\epsilon} \\ &=-\frac{1}{2} + \lim\limits{\epsilon \downarrow 0}\frac{1}{2\epsilon^{2}}\\ &= \infty \end{align*} $$ Thus, by definition of Extended L1 Functions, $X \not\in \bar{L^{1}}$. $\blacksquare$