Let $f $ be a continuous function such that
$$ \left| \int_{0}^{1} \phi(x) f(x)~dx \right| \le C_{\phi}$$
for every $\phi \in \mathcal{D}(0,1)$. Is it true that
$$ \left| \int_{K} f(x)~dx \right| \le C $$ for every compact $K \subset [0,1]$? In other words can we conclude that $f \in L^{1}_{loc}(0,1)$?
I think it could be true by the arbitrary-ness of the test function but I am having trouble constructing an argument.
Here is a proof. The assumption is: for all $\phi \in C_c^\infty(0,1)$ with $\phi\ge0$ the integral $\int_0^1 f(x) \phi(x)dx$ exists in the Lebesgue sense and is finite.
First, assume $f\ge0$. Let $K\subset(0,1)$ be compact. Take non-negative $\phi\in C_c^\infty(0,1)$ such that $\phi=1$ on $K$ (this can be achieved by mollification). Then $$ \int_K f(x) dx \le \int_0^1 f(x) \phi(x) dx < + \infty, $$ and $f$ is locally integrable.
Second, consider the general case. By definition of the Lebesgue integral and by assumption, we have that the integrals of the positive and negative part of $f\cdot\phi$ exist and are finite. Then $\int_0^1 (f(x))^+ \phi(x) dx < \infty$ for all smooth and non-negative $\phi$ implies local integrability of $f^+$ by the first part of the proof. Similarly, $f^-$ is locally integrable, which implies local integrability of $f$.