Definition (Local Integrability) $:$ Let $\Omega \subseteq \mathbb R^N$ be an open subset and $f : \Omega \longrightarrow \mathbb R$ be a real valued function defined on $\Omega.$ Then $f$ is said to be locally integrable on $\Omega$ if for any compact subset $K \subseteq \Omega$ we have $f \in L^1 (K).$ The set of all locally integrable functions on $\Omega$ is denoted by $L^1_{\text {loc}} (\Omega).$
Proposition $:$ Let $f \in L^1_{\text {loc}} (\Omega)$ be such that $$\int_{\Omega} f g\ dx = 0$$ for all $g \in C_c (\Omega).$ Then $f = 0$ almost everywhere on $\Omega.$
The proof of this theorem is divided into parts.
The author* first proved the proposition for integrable function $f$ defined on $\Omega$ by using the density of $C_c (\Omega)$ in $L^1 (\Omega),$ where the Lebesgue measure $|\Omega|$ of $\Omega$ is taken to be finite. I have no issue with this part.
In the next step he proved the proposition for any locally integrable function $f$ on $\Omega$ for any open subset $\Omega \subseteq \mathbb R^N.$ For that he restricted $f$ to $\Omega_n$ where $\Omega_n = \Omega \cap B (0, n).$ Now since $\overline {\Omega_n}$ is a compact subset of $\Omega$ and $f$ is locally integrable on $\Omega$ it follows that $f \in L^1 \left (\overline {\Omega_n} \right )$ and hence $f \in L^1 (\Omega_n).$ Now he said that since $\left |\Omega_n \right | \lt +\infty$ we can use the first part to conclude that $f = 0$ almost everywhere on $\Omega_n$ for all $n \geq 1$ and since $\Omega = \bigcup\limits_{n = 1}^{\infty} \Omega_n$ it follows that $f = 0$ almost everywhere on $\Omega,$ as countable union of sets of measure zero is again a set of measure zero. This completes the proof.
For the second part how did the author know that $\int_{\Omega_n} f h\ dx = 0$ for all $h \in C_c (\Omega_n)$ eventually after a certain stage by the given condition? This is where I got stuck and couldn't able to understand the reason behind it. Could anyone please make it clear to me?
Thanks a bunch.
*Source $:$ Functional Analysis by S. Kesavan (Proposition 6.3.3, Page no. $173$).
Best Answer
By my accounting, the proof has only one error, right here:
As defined, $\overline {\Omega_n}$, while compact, may not be a subset of $\Omega$, so we would only be able to conclude $f \in L^1_\text{loc}(\Omega_n)$, which isn't sufficient for the rest of the proof.
The simplest fix I can see is to write instead $$\Omega_n = \Omega \cap B(0,n) \setminus \overline{\bigcup_{x \not\in \Omega} B(x,\tfrac1n)}$$
From here, we can conclude $f \in L^1(\overline{\Omega_n})$ for the same reasoning given in the original argument, but now there is something to show in order to conclude $\Omega = \bigcup_n \Omega_n$.
As for your question, I don't actually see your issue. Let's run through the details a little more closely in the proof.
We are assuming that $f \in L^1_{\text{loc}}(\Omega)$ and that $\int_{\Omega} fg\,\rm{d}x = 0$ for every $g \in C_c(\Omega)$.
With the above fix, $f|_{\Omega_n} \in L^1(\Omega_n)$ and $$\int_{\Omega_n}f|_{\Omega_n}g\,\rm{d}x = \int_\Omega fg\,\rm{d}x = 0$$ for every $g \in C_c(\Omega_n) \subseteq C_c(\Omega)$. By the first part, $f|_{\Omega_n} = 0$ almost everywhere on $\Omega_n$.
Now use the fact, as mentioned, that $\Omega = \bigcup_n \Omega_n$ and the countable union of measure zero sets has measure zero.
Hopefully that clears it up.