Suppose $\ f:\ X\rightarrow(0,\infty)$ is a measurable function.
If $$\int_{X} f\ d\mu<\infty\ $$ and $$\int_{X} \dfrac1f\ d\mu<\infty $$ Show that $\mu(X)<\infty$.
Suppose $\ f:\ X\rightarrow(0,\infty)$ is a measurable function.
If $$\int_{X} f\ d\mu<\infty\ $$ and $$\int_{X} \dfrac1f\ d\mu<\infty $$ Show that $\mu(X)<\infty$.
Daniel Fischer's hint is right on target: $$2\mu(X)\le \int_X (f+f^{-1}) <\infty$$ Or, if one feels like using a (mis-)named result, the Cauchy-Schwarz inequality yields $$\mu(X) =\int_X f^{1/2}f^{-1/2} \le \sqrt{\int_X f}\sqrt{\int_X f^{-1}}<\infty$$