Convergence of logarithm of $L^1$ functions in $L^1$.

Question

Let $R$ be a probability measure, and $L^1(R) := \{ \phi : \int |\phi| ~dR < \infty\}$. Let $\phi_n$ be a convergent sequence of functions in $L^1(R)$ converging to $\phi \in L^1(R)$. That is, as $n \to \infty$, $$ \int |\phi_n - \phi| ~dR \to 0 .$$ If we know that $\log (\phi_n) \in L^1(R)$ and $\log (\phi) \in L^1(R)$ then will $$ \int |\log(\phi_n) - \log(\phi)| ~dR \to 0 $$ as $n \to \infty$? Intutively I can see that it will hold true but I am not able to come up with a mathematical argument for the same. Please do help.

Answer 1

Let $\phi_n (x)=e^{-n}$ if $0<x<\frac 1 n$ and $1$ on $(\frac 1 n ,1)$. (on the interval $(0,1)$ with Lebesgue measure). Then $\phi_n \to 1$ in $L^{1}$ but $\log \phi _n$ does not converge to $0$ in $L^{1}$. [I have used the standard definition of $L^{1}(R)$. If you are keen on defining $\phi_n$'s and $\phi$ on the whole line simply extend the definitions by making all equal to a fixed integrable function, say $e^{-|x|}$ outside $(0,1)$].