why is the Laplace transform of local integrable function with support on $[0,\infty)$ analytic?

There is a proposition about Laplace transform, but I don't know how to prove it.

Let $$f \in L^1_{loc}(\mathbb{R})$$, $$\operatorname{supp}(f) \subset[0, \infty)$$, such that $$a$$ is the abscissa of absolute convergence of the Laplace transform $$F$$ of $$f$$. Then $$z \mapsto F(z)$$ is analytic in the half plane $$\operatorname{Re}(z)>a$$.

A point $$a$$ is called the abscissa of absolute convergence of the Laplace transform $$F$$ of $$f$$, if $$a$$ is the minimum real number such that $$\int_0^\infty |f(t)| e^{-Re(z)t} dt$$ exists for any $$z$$ with $$Re(z) > a$$.

Morera's theorem and Fubini's theorem provide a quick solution. Let $$\gamma:[0,1]\to \{z:\Re(z)>a\}$$ be a piecewise $$C^1$$ closed curve. Then, let $$\theta_*$$ be such that $$a<\Re(\gamma(\theta_*)=\min_{0\leq \theta\leq 1} \Re(\gamma(\theta))$$, then $$|e^{-\gamma(\theta)t}|\leq e^{-Re(\gamma(\theta_*)t}$$ and $$\int_0^1\int_0^\infty |e^{-\gamma(\theta)t}||\gamma'(\theta)||f(t)|\mathrm dt\mathrm d\theta\leq \int_0^1\int_0^\infty e^{-\Re(\gamma(\theta_*))t}|\gamma'(\theta)||f(t)|\mathrm dt\mathrm d\theta<\infty$$ so that by Fubini-Tonelli, $$\int_\gamma F(z)\mathrm dz=\int_0^\infty f(t)\int_\gamma e^{-zt}\mathrm dz\mathrm dt=0$$ and $$F$$ is holomorphic by Morera's theorem.