In a finite measure space, does it follow that $f\in L^1 \implies f\in L^2$? Clearly on $\{|f|\le 1\}$ (which is measurable as long as $f$ is measureable) we must have $f^2\le |f|$ so that $f^2$ is integrable (regardless of finiteness of measure). But does it follow that $f^2$ is also integrable outside this set?
I guess it is, because if the measure is finite measure of whole set must be finite, and to be non-integrable $f^2$ would have to be unbounded in a set with nonzero measure. But this is not possible with $f$ being integrable. But I can't formulate it in rigourous way (or in fact this claim is false?).
Any helps appreciated.
 A: No. $\frac{1}{\sqrt{x}}\in L^1 (0,1) \wedge \not\in L^2(0,1)$.
A: Consider $([0,1], \mathcal{B}[0,1], \lambda)$ where $\lambda$ is the Lebesgue measure.  Let $f: [0,1] \to \mathbb{R}$ be defined by
$$
f(x) := \frac{1}{x^\alpha}, \qquad \alpha > 0.
$$
Then for $p \geq 1$ (just so $|| \cdot ||_p$ is a norm),
\begin{align*}
||f||_p^p & = \int |f|^p\, d\lambda = \int \left(\frac{1}{x^\alpha}\right)^p \, d\lambda = \int x^{-\alpha p} \, d\lambda = \frac{1}{1 - \alpha p} < \infty \\
& \iff p < \frac{1}{\alpha}.
\end{align*}
For avid19's example, $\alpha = \frac{1}{2}$ and so $f \in \mathcal{L}^p$ for $1 \leq p < 2$, but not in $L^p$ otherwise.
This is actually a good rule of thumb: for function which has a "blow up problem" (as $\frac{1}{\sqrt{x}}$ near the origin), larger $p$'s are worse.
Update I should add your result holds for the counting measure, in which the Lebesgue spaces are usually denoted $\mathcal{l}^p$.  That is, $\mathcal{l}^p \subset \mathcal{l}^q$ for any $0 < p < q \leq \infty$, and $||f||_q \leq ||f||_p$.
