The following problem is from Ch. 15 of Spivak's Calculus. I asked a separate question about understanding the solution manual solution to part $(c)$. Since my solution differed from that solution, I would like to know if it is correct.
*26. It is an excellent test of intuition to predict the value of
$$\lim\limits_{\lambda\to \infty} \int_a^b f(x)\sin{(\lambda x)}dx$$
Continuous functions should be the most accessible to intuition, but once you get the right idea for a proof the limit can easily be established for any integrable $f$
(a) Show that $\lim\limits_{\lambda\to \infty} \int_a^b \sin{(\lambda x)}dx=0$, by computing the integral explicitly.
(b) Show that if $s$ is a step function on $[a,b]$, then $\lim\limits_{\lambda\to \infty} \int_a^b s(x)\sin{(\lambda x)}dx=0$.
(c) Finally, use Problem 13-26 to show that $\lim\limits_{\lambda\to \infty} \int_a^b f(x)\sin{(\lambda x)}dx=0$ for any function $f$ which is integrable on $[a,b]$. This result, like Problem 12, plays an important role in the theory of Fourier series; it is known as the Riemann-Lebesgue Lemma.
Here is my attempt at a solution to part $(c)$ (I show solutions to parts $(a)$ and $(b)$ in another question about this problem)
Assume $f$ is integrable on $[a,b]$.
Given a number $\lambda$, let $P_{\lambda}=\{t_0,...,t_{n_{\lambda}}\}$ be a partition of $[a,b]$ such that for any $i=1,...,n_{\lambda}$ we have that in the partition subinterval $[t_{i-1},t_i]$, $\sin{(\lambda x)}$ is either always $\geq 0$ or $< 0$.
Let $s_1$ and $s_2$ be step functions for $P_{\lambda}$ such that $s_1<f<s_2$.
Then, for any partition subinterval $[t_{i-1},t_i]$, we have one of the two following possibilities
$\sin{(\lambda x)}\geq 0 \implies s_{1,i}\sin{(\lambda x)}<f(x)\sin{(\lambda x)}<f(x)\sin{(\lambda x)}<s_{2,i}\sin{(\lambda x)}\tag{1}$
$\sin{(\lambda x)}< 0 \implies s_{2,i}\sin{(\lambda x)}<f(x)\sin{(\lambda x)}<f(x)\sin{(\lambda x)}<s_{1,i}\sin{(\lambda x)}\tag{2}$
In each case, integrate
$s_{1,i} \int_{t_{i-1}}^{t_i}\sin{(\lambda x)}<\int_{t_{i-1}}^{t_i}f(x)\sin{(\lambda x)} < s_{2,i} \int_{t_{i-1}}^{t_i}\sin{(\lambda x)}=0\tag{3}$
$s_{2,i} \int_{t_{i-1}}^{t_i}\sin{(\lambda x)}<\int_{t_{i-1}}^{t_i}f(x)\sin{(\lambda x)} < s_{1,i} \int_{t_{i-1}}^{t_i}\sin{(\lambda x)}=0\tag{4}$
and take the limit
$0=\lim\limits_{\lambda\to \infty} s_{1,i} \int_{t_{i-1}}^{t_i}\sin{(\lambda x)}<\lim\limits_{\lambda\to \infty} \int_{t_{i-1}}^{t_i}f(x)\sin{(\lambda x)} <\lim\limits_{\lambda\to \infty} s_{2,i} \int_{t_{i-1}}^{t_i}\sin{(\lambda x)}=0\tag{5}$
$0=\lim\limits_{\lambda\to \infty} s_{2,i} \int_{t_{i-1}}^{t_i}\sin{(\lambda x)}<\lim\limits_{\lambda\to \infty} \int_{t_{i-1}}^{t_i}f(x)\sin{(\lambda x)} <\lim\limits_{\lambda\to \infty} s_{1,i} \int_{t_{i-1}}^{t_i}\sin{(\lambda x)}=0\tag{6}$
Thus,
$$\lim\limits_{\lambda\to \infty} \int_{t_{i-1}}^{t_i}f(x)\sin{(\lambda x)}=0\tag{7}$$
And since
$$\int_a^bf(x)\sin{(\lambda x)}=\sum\limits_{i=1}^{n_{\lambda}} \int_{t_{i-1}}^{t_i}f(x)\sin{(\lambda x)}$$
Then
$$\lim\limits_{\lambda\to \infty}\int_a^bf(x)\sin{(\lambda x)}$$
$$=\lim\limits_{\lambda\to \infty}\sum\limits_{i=1}^{n_{\lambda}} \int_{t_{i-1}}^{t_i}f(x)\sin{(\lambda x)}\tag{8}$$
$$=0$$
where we are using the result from $(7)$.
Now, $(8)$ looks slightly weird. As $\lambda\to\infty$, the number of partition subintervals increases (ie, $n_{\lambda}\to\infty$), because $\sin{(\lambda x)}$ gets compressed. However, I think we can always form the partition as I defined it.
Is this solution correct?
