Let $X=<0,1>$, take the borel sigma algebra, and the lebesgue measure. Consider $g(x)=\dfrac{1}{x^{\frac{1}{2}}}$. Show that $g\in L_p$ iff $1\leq p<2$.
I have done this:
$\int_{<0,1>}\mid g\mid ^p d\mu = \mid x\mid ^{-\dfrac{p}{2}}.\mu(<0,1>)=\mid x\mid ^{-\dfrac{p}{2}}$. But why this is not $\infty$ iff $p<2$?
I'll appreciate any help, thanks.
The Lebesgue integral of a function should be a (extended) real number, not a function; so the statement
$$\int_0^1 |g|^p d\mu = |x|^{-p/2} \mu([0,1])$$
is not correct.
One way to proceed here is to reduce this to an integral that can be computed in the classical way using a Riemann integral: If we have a continuous (or almost continuous) function that's bounded, we can just integrate like in ordinary calculus. Now notice that by the Monotone Convergence Theorem,
\begin{align*} \int_0^1 |g|^p d\mu &= \lim_{\epsilon \to 0^+} \int_{\epsilon}^1 |g|^p d\mu \\ &= \lim_{\epsilon \to 0^+} \int_{\epsilon}^1 x^{-p/2} d\mu \\ &= \lim_{\epsilon \to 0^+} \frac{x^{1 - p/2}}{1 - p/2} \Big|_{\epsilon}^1 \end{align*}
and this limit can be computed directly (namely, the relevant cases to study are based on the sign of $1 - p/2$, or how $p$ compares to $2$).