$\sup\limits_{\phi} \int_{[0,1] } \log \phi = \int_{[0,1]} \log f$

Question

Let's say we have a measurable function $f:[0,1] \rightarrow (0, \infty)$. Approximate $f$ from below by a simple function $\phi$, with $\phi(x) > 0$ for all $x$. Then $$\int f = \sup\limits_{\phi} \int \phi$$ My question: is it also true that $$\sup\limits_{\phi} \int \ln \phi = \int \ln f\space?$$ I believe this is at least true when $f(x) > 1$ for all $x$ (in which case we can replace the condition $\phi(x) > 0$ with $\phi(x) > 1$). In this case, $\log f > 0$, and $\int \log f$ is the supremum of all integrals $\int \psi$, where $\psi$ is a positive valued simple function which $\leq \log f$. But the set of all such $\psi$ coincides exactly with the set of $\log \phi$, where $\phi$ is a simple function with $1 \leq \phi \leq f$. This is because any such $\phi$ is equal to $$\sum\limits a_i \chi E_i$$ with $a_i = \log e^{a_i}$.