Consider the following fragment from Murphy's book "$C^*$-algebras and operator theory" (p96: Toeplitz operators):
Question:
Why do we have $$\forall g \in C(\Bbb{T}): \int fg \,d \lambda = 0 \implies f \, d \lambda = 0$$
Maybe we should use that $C(\Bbb{T})$ is dense in $L^1(\Bbb{T})$ together with DCT?
The map $$F: (L^\infty(\mathbb{T}, |f \, d\lambda|), \|\cdot\|_1) \to \mathbb{C}, \, g \mapsto \int_{\mathbb{T}} fg \, d\lambda$$ is a bounded linear functional and hence continuous since $$|F(g)| = \left|\int_{\mathbb{T}} fg \, d\lambda \right| = \left|\int_{\mathbb{T}} g \, d(f\, d\lambda)\right| \leq \int_{\mathbb{T}} |g| \, d |f \, d\lambda| = \|g\|_1$$ where we used that for any bounded measurable $g: \mathbb{T} \to \mathbb{C}$ we have $$\int_{\mathbb{T}} fg \, d\lambda = \int_{\mathbb{T}} g \, d(f \, d\lambda).$$ Furthermore $C(\mathbb{T})$ is $\|\cdot\|_1$-dense in $L^\infty(\mathbb{T}, |f \, d\lambda|)$ and $F$ vanishes on $C(\mathbb{T})$, so we obtain $F = 0$. Now $g = \frac{\overline{f}}{|f|+1} \in L^\infty(\mathbb{T}, |f \, d\lambda|)$ and therefore $$0 = F(g) = \int_{\mathbb{T}} \frac{|f|^2}{|f|+1} \, d\lambda,$$ so $f = 0$ $\lambda$-almost everywhere.
More generally it can be shown that for a regular complex measure $\nu$ on a locally compact Hausdorff space $X$ we have $$\int_X \phi \, d\nu = 0 \ \ \forall \phi \in C_c(X) \implies \nu = 0.$$ This is one component of the complex Riesz-Markov-Kakutani representation theorem, but the proof is analogous to the above since $C_c(X)$ is dense in $L^1(X, |\nu|)$.