I couldn't find the question on the website.
I need to determine the integral $\lim_{n\rightarrow \infty }\int_{\left[ 0,1\right] }f_{n}d\mu $, if it exists, where $\mu $ is the Lebesgue measure and $$ f_{n}=\frac{\sin \left( \left( x+\frac{1}{n}\right) ^{2}\right) -\sin \left( \left( x-\frac{1}{n}\right) ^{2}\right) }{\sin \frac{1}{n}} $$ Now I can see that $$\frac{\sin \left( \left( x+\frac{1}{n}\right) ^{2}\right) -\sin \left( \left( x-\frac{1}{n}\right) ^{2}\right) }{\sin \frac{1}{n}}=\frac{\sin \left( \left( x+\frac{1}{n}\right) ^{2}\right) -\sin \left( \left( x-\frac{1}{n}\right) ^{2}\right) }{\frac{2}{n}}2\frac{\frac{1}{n}}{\sin \frac{1}{n}}$$ and using the fact that $g^{\prime }\left( x\right) =\lim_{h\rightarrow 0}\frac{g\left( x+h\right) -g\left( x-h\right) }{2h}$ for $g\left( x\right) =\sin \left( x^{2}\right) $ and that $\frac{\frac{1}{n}}{\sin \frac{1}{n}}\longrightarrow 1$, we conclude that $f_{n}\longrightarrow 4x\cos x^{2}$ but I am having trouble moving the limit under the integral sign. What I think is that, since this sequence of functions converges pointwise, they must be bounded by a finite function $% m\left( x\right) $, which is Lebesgue integrable and then we can apply the DCT. Is this okay? I tried to show that convergence is uniform but my hand wouldn't move.
In general, pointwise convergence of $f_n$ does not imply the existence of dominating functions. But in this case it is easy.
Note
\begin{align} \frac{g(x+h) - g(x-h)}{2h} &= \frac 12 \frac{g(x+h) - g(x)}{h} + \frac 12 \frac{g(x) - g(x-h)}{h}\\ &= \frac{1}{2} (g'(x_1) +g'(x_2)) \end{align}
by the Mean Value Theorem. Since $|g'|\le 2$ on $[0,1]$, we see that $|f_n|\le 8$ when $n$ is large enough. Thus you can apply Dominate Convergence Theorem.