# $L^1 \subseteq L^2$?

Let $(\Omega, \Sigma, \mu)$ - be a measure space. Suppose, $\mathbb{E}(X) < \infty$, i.e. $X \in L^1(\Omega, \Sigma, \mu)$.

Given this, it follows $\mathrm{ess}\,\sup \lvert X \rvert < \infty$, otherwise $\mathbb{E}(X^+)$ or $\mathbb{E}(X^-)$ would be $\infty$. So we can deduce $X \in L^{\infty}(\Omega, \Sigma, \mu)$.

By Hölder's Inequality it follows

$$\lvert\lvert X\cdot X \rvert\rvert_1 \leq \lvert\lvert X \rvert\rvert_1 \lvert\lvert X \rvert\rvert_{\infty} < \infty$$

which means $\mathbb{E}(X^2) = \lvert\lvert X \rvert\rvert_2^2 < \infty$

I feel like this should not be true.

Where is the above reasoning wrong?

Unfortunately, the above reasoning is indeed flawed; you cannot conclude $$\operatorname{ess sup}|X| < \infty,$$ because even though $|X|$ can attain $\infty$ only on a nullset, it can still become arbitrarily large on non-nullsets.
• Concrete example: $1/\sqrt{x}$ on $(0,1)$. – Wojowu Feb 15 '18 at 9:11