Let $f_t(x) = \frac{\sin tx}x$ and $\varepsilon>0$. Further, consider a test function $\varphi \in C_0^\infty(\mathbb R)$ with $\operatorname{supp} \varphi \subseteq [-R, R]$ for some $R>0.$
I have proved that $$\displaystyle \int_{\mathbb R\setminus[-\varepsilon, \varepsilon]} f_t(x) \; \mathrm dx \to 0$$ when considering the limit $t\to \infty$. I want to prove $$\displaystyle \int_{\mathbb R\setminus[-\varepsilon, \varepsilon]} f_t(x) \varphi(x) \; \mathrm dx \to 0$$ from this.
The problem is that $f_t \not\to 0$, so dominated convergence doesn't work. Moreover; $f_t$ is not absolutely integrable and, even more so, the integral $\displaystyle\int_{\mathbb R} f_t(x) \; \mathrm dx$ only exists in a generalized Riemann sense (with an undefined Lebesgue integral).
Can I prove this without using Riemann-Lebesgue's lemma (as they do here), if the limit $\displaystyle \int_{\mathbb R\setminus[-\varepsilon, \varepsilon]} f_t(x) \; \mathrm dx \to 0$ is known?
If it is helpful we can also assume that $\displaystyle\int_\Omega f_t(x) \; \mathrm dx \to 0$ for any closed subset $\Omega \subset \mathbb R$ with $0 \not\in \Omega$, as in the case for $f_t(x) = \frac{\sin tx}x$.
You can get the result integrating by parts. For $\varepsilon<R$ $$ \int_{\mathbb R\setminus[-\varepsilon, \varepsilon]} f_t(x)\, \varphi(x) \, {\mathrm d}x =\int_{-R}^{-\varepsilon}f_t(x)\, \varphi(x)\,\mathrm dx+\int_ {\varepsilon}^Rf_t(x\,) \varphi(x)\,\mathrm dx. $$ Now \begin{align} \int_ {\varepsilon}^Rf_t(x)\, \varphi(x)\,\mathrm dx&=\int_{\varepsilon}^R\frac{\varphi(x)}{x}\,\sin(t\,x)\,\mathrm dx\\ &=-\frac{\varphi(x)}{x}\,\frac{\cos(t\,x)}{t}\Bigr|_\varepsilon^R+\int_{\varepsilon}^R\Bigl(\frac{\varphi(x)}{x}\Bigr)'\frac{\cos(t\,x)}{t}\,\mathrm dx. \end{align}