$f^2 $ is Lebesgue Integrable, imply $f$ is Lebesgue integrable? Suppose $f^2 $ is Lebesgue Integrable. Will it imply $f$ is Lebesgue integrable.
This is not true for Riemann integrable case. I have example, but that is not working for here. I guess that it is true, some one help me please. 
 A: Suppose that $S \subset [0,1]$ is a non measurable real subset. Then the function $f$ defined by
$$\begin{array}{l|rcl}
f : & \mathbb R & \longrightarrow & \mathbb R \\
    & x & \longmapsto & -1 \text{ if } x \in S\\\
    & x & \longmapsto & 1 \text{ if } x \in [0,1] \setminus S\\\
    & x & \longmapsto & 0 \text{ else}\\
 \end{array}$$
has square $f^2$ which is measurable and integrable. However, $f$ is not measurable, hence not integrable.
So an interesing additional question: were you referring to a measurable function?
A: To fix notation, suppose
$$ \int_A f^2 \, d\mu < \infty, $$
and suppose $f$ is measurable (see the other answers for the non-measurable case).
The answer depends on the measure of $A$, because the Cauchy-Schwarz inequality says
$$ \int_A f \cdot 1 \, d\mu \leqslant \left( \int_A f^2 \, d\mu \right)\left( \int_A \, d\mu \right) = \mu(A) \int_A f^2 \, d\mu. $$
Therefore it is true if $\mu(A)$ is finite. If $\mu(A)$ is infinite, it is false: $1/x$ on $[1,\infty)$ with Lebesgue measure, for example.
A: It's true on $]a,b[$ but false on $\mathbb R$. For example $x\mapsto \frac{1}{x}$ is $L^2(1,\infty )$ but not $L^1(1,\infty )$.
For the case $]a,b[$, you have by Cauchy-Schwarz that $$\int_a^b |f|\leq \sqrt{b-a}\sqrt{\int_a^b|f|^2}$$
what prove the claim.
A: Lets assume $f$ is measurable. On a finite measure space, it does because
$$ ∫_X f = ∫_X f\times 1 \overset{C-S}{\leq} ∫_Xf^2∫_X 1 < ∞ $$
It isn't true on say $ℝ$, since we can take $f(x) = \frac{\Bbb 1_{|x|>1}}{x}$.
A: We could try $$f(x) = \frac{1}{|x|}$$ for $|x|>1$ and $$f(x) = 1$$ for $|x| \le 1$. It's easy to see that this is not integrable where as its square is.
