Let $(X, \mathcal{A}, \mu)$ be a measurable space that has a function $f: X \rightarrow [0, +\infty]$ such that $$\int_Xfd\mu < \infty $$ and $$\int_Xf^{-1}d\mu < \infty $$Show that $\mu(X) < \infty$.
In order to show this, I decided to use the Holder's inequality. Let $p,q \in \mathbb{N}^*$ such that $\frac{1}{p} + \frac{1}{q} = 1$
Then we have $$\int_X|f \times f^{-1}|d\mu \leq (\int_X f^pd\mu)^{\frac{1}{p}}(\int_X (f^{-1})^{q}d\mu)^{\frac{1}{q}}< \infty$$
As $$\int_X|f \times f^{-1}|d\mu = \int_Xd\mu = \mu(X)$$ we thus have $\mu(X) < \infty$
Is my reasoning correct? I am not too used to the usage of Holder's inequality in measurable spaces.
$$y + \frac{1}{y} \geq 1 \qquad \text{for all $y \geq 0$}$$
and so
$$f(x)+ \frac{1}{f(x)} \geq 1 \qquad \text{for all $x \in X$}.$$ Integrate both sides...