I want to calculate the limit of following Lebesgue-integral:
$$ \lim_{n \rightarrow \infty} \int_{[0, \infty)} \frac{\sin(e^x) }{1+nx^2}\,\mathrm dx$$
Therefore I wanted to apply Lebesgue's dominated convergence theorem. $f_n(x)$ is measurable and $ f_n \rightarrow 0$ pointwise. Now it holds:
$$ \left|\frac{\sin(e^x) }{1+nx^2}\right| \leq \frac{1}{1+x^2} :=g(x) $$
The improper integral over does converge. That means f is lebesgue integrable. Therefore $$ \lim_{n \rightarrow \infty} \int_{[0, \infty)} \frac{\sin(e^x) }{1+nx^2}\,\mathrm dx = \int_{[0, \infty)} \lim_{n \rightarrow \infty} \frac{\sin(e^x) }{1+nx^2}\,\mathrm dx =0$$ Consider $ f_n(0) = sin 1$ does not converge to $ 0$. So I can't apply the theorem, can I ?