I am reading Banach Spaces of Analytic Functions by Hoffman. In Page 52, the authors claim this inequality as passing remark:
Let $f \in L^1 (\mathbb T)$ and $f(e^{i0})\ne 0$ and suppose that $$\frac{1}{2\pi}\int \log (|f| +\varepsilon) d\theta \ge \log |f(e^{i0})| \qquad \forall \varepsilon >0$$ We can then show that: $$\frac{1}{2\pi}\int \log (|f|) d\theta \ge \log |f(e^{i0})|$$
I initially thought of using Fatou's inequality to prove this but I do not think that will work here. Is there any other way passing the limit inside the integral here?
EDIT
I think I have found the proof but I would like some verification.
Note that:
$$1 = \log e \le\log (|f| + e) \le |f| +e$$ Hence, $\log (|f| + e)$ is integrable. We have that $\log(|f| + 1/n) \le \log (|f|+e)$ for every $n \in \mathbb N$ by assumption. Thus, we have that by Dominated Convergence theorem that $$\int \log(|f|) \frac{d\theta}{2\pi}= \lim_{n\to \infty} \int \log(|f| + 1/n) \frac{d\theta}{2\pi}$$