I came across this old qualifying exam problem and I am stuck. Suppose $\mu$ is a probability measure and $f:X\rightarrow [0,\infty]$ is measurable. Suppose that $$G=\displaystyle{e^{\int \log(f) d \mu}}$$ is finite. Then show that $G$ is defined (that is, that $\int \log(f) d \mu$ is defined) and show that $\lim_{t\rightarrow _{0^+}}(\int f^td\mu)^{\frac{1}{t}}=G$.
I rewrote $(\int f^td\mu)^{\frac{1}{t}}$ as $e^{1/t\ln(\int f^td\mu)}$, and I know that $\ln(\int f^td\mu)\leq(\int f^td\mu)-1$ by concavity of $\log$. But, I am getting stuck after this, and also, on how to show $G$ is defined.
