False proof that every continuous function is holomorphic

Let $\Omega$ be an open subset of $\mathbb{C}$ and $f:\Omega\to\mathbb{C}$ be a continuous function. Consider the following function: $$F(z)=\int_{[z_0,z]}f(w)\:\mathrm{d}w,$$ where $z_0$ is a fixed complex number.

Firstly I will prove that $F$ is holomorphic.

$$\lim_{h\to 0} \frac{F(z+h)-F(z)}{h} = \lim_{h\to 0} \frac{1}{h}\left(\int_{[z_0,z+h]} f(w)\:\mathrm{d}w - \int_{[z_0,z]} f(w)\:\mathrm{d}w\right)= \lim_{h\to 0} \frac{1}{h}\int_{[z,z+h]} f(w)\:\mathrm{d}w= \lim_{h\to 0} \int_0^1 f(z+th)\:\mathrm{d}t= \int_0^1 \lim_{h\to 0}f(z+th)\:\mathrm{d}t= f(z)$$

We can pass the limit under the integral sign by the following reason: $[0,1]$ is compact and $f$ is continuous. Hence there exists a real number $M>0$ such that $|f(z+th)|\leq M$ for all $t\in[0,1]$. This means that we can use the continuity under the (Lebesgue's) integral sign theorem.

Since $F$ is holomorphic and holomorphic functions have derivatives of all orders, $F'=f$ also has derivatives of all orders. In particular, $f$ is holomorphic.

Ok, this is clearly an absurd. However I don't know where is my error.

Solution 1:

The problem is two-fold. First, $F$ may be not well defined. Even if it is, the step where you replace the integral over $[z_0,z+h]$ minus $[z_0,z]$ by that over $[z,z+h]$, you are assuming $f$ integrates over $0$ over triangles. This condition is known to be equivalent (locally), to being holomorphic. In this case, you are more or less replicating the proof that over convex sets (or more generally star shaped sets), functions that integrate over zero over triangles are integrable.