Let $f(x)$ be a continuous complex-valued function on $[0,1]$ and define $$g(z)=\int_0^1\frac{f(x)}{x-z}\,dx.$$ Where is $g(z)$ defined?
It is obvious that $g(z)$ is defined on $\mathbb{C}\setminus[0,1]$. But I am wondering about points in $[0,1]$. Is it true that $g(z)$ is defined on any point $w\in[0,1]$ such that $f(w)=0$? I am trying to use the fact that $f(x)$ is uniformly continuous on $[0,1]$, but I cannot deduce any information about whether $\frac{f(x)}{x-z}$ is continuous at $w$.
Let's take $w = 0$. If $f(0) = 0$, can we conclude that $g(0)$ is defined?
$$g(0) = \int_{0}^{1} \frac{f(x)}{x} dx$$
The $x$ in the denominator desires to make the integral diverge by introducing a singularity of order $1$. If $f(x)$ vanishes at zero like $x^{\alpha}$ for $\alpha > 0$ then the whole integrand has a singularity of order $1 - \alpha$, which converges. So if the integral is going to still diverge, the zero of $f(x)$ at $0$ better be really subtle. Perhaps
$$f(x) = \frac{1}{\log (x/2)}$$
This is continuous on $[0,1]$ because $\log(x/2) = 0$ at $x = 2$. Can you evaluate the integral yourself? This should only require basic calculus tricks.