Lebesgue integrable implies bounded a.e

Question

I am thinking on the question: does Lesbesgue integrable implies bounded a.e? I think it does. Is Chebyshev the correct way to show it?

Answer 1

No, Chebyshev's theorem only shows that $f$ is finite a.e. $$ m(\{f = \infty\}) = m(\bigcap_n^\infty \{ f \geq n \}) = \lim_{n \to \infty} m(\{f \geq n\}) \leq \lim_{n \to \infty} \dfrac{1}{n} \int f = 0 )$$

Now take $f = \dfrac{1}{\sqrt{x}}$, $f \in L_1(0,1)$ but is not bounded a.e.

Also remember that if $m(X) < \infty$, $L_{\infty} \subseteq L_p$ for all $p \geq 1$, and $L_{\infty}$ are the ones bounded a.e.

Answer 2

No,it's not true. Because then you would have that $L^{1}(X)\subset L^{\infty}(X)$ with norm $\|.\|_{\infty}$.

$L^{1}(X)$ with $\|.\|_{\infty}$ is closed in $L^{\infty}(X)$ and thus it's complete. Wrong!