Suppose $f\_n$ is a sequence of holomorphic functions taking $\mathbb{D} \to \mathbb{C}$ where $\mathbb{D}$ is the unit disk. Further, $f\_n$ has a continuous extension to $\overline{\mathbb{D}}$. We can assume $\sum\_{n=0}^\infty f\_n$ converges normally on compact subsets of $\mathbb{D}$ to a holomorphic function $f$.


Additionally, for each $f\_n$ we know $\int\_C |f\_n| < \infty$ for any contour $C \subset \overline{\mathbb{D}}$. Less strictly, so we don't have enough to use the monotone convergence theorem, we know that 
$\sum\_{n=0}^\infty |\int\_C f\_n| < \infty$. But additionally, for any closed contour $C^\*$ in $\overline{\mathbb{D}}$ we know $\int\_{C^\*}f\_n = 0$.


Must it follow


$$\int\_C f = \sum\_{n=0}^\infty \int\_C f\_n$$


I'm asking because all the naive instances where the monotone convergence theorem fail are exempt from these criterion. I think there's something more subtle in this instance.


I've been able to strengthen the condition to a proof that


$$\int\_C f = \sum\_{n=0}^\infty \int\_C f\_n \,\,\Leftrightarrow\,\, \sum\_{n=0}^\infty \sup\_{C\subset \overline{\mathbb{D}}}|\int\_Cf\_n| < \infty$$


Or rephrase it to


$$\sum\_{n=0}^\infty |\int\_C f\_n| < \infty\,\, \Leftrightarrow \,\,\,f \, \text{can be continuously extended to}\,\overline{\mathbb{D}}$$


Those are the two avenues I've taken. None have really given me an answer.


Any suggestions, comments, questions are greatly appreciated.



Interchanging sums and integrals in a specific instance
Take $f\_n(z) = 3z^{3n} - z^n$ and $C = \{ e^{i\theta} \ | \ \theta \in (0,\pi) \}$. It satisfies all of your hypotheses with 
$$
f(z) = \frac{2+z}{1+z+z^2}
$$
and $\int\_C f\_n =0$ for $n \geq 1$. But $\int\_C f$ is not a convergent integral.


Of course the stronger condition $\sum\_n \int\_C |f\_n| < \infty$ is sufficient for your conclusion to hold.


