The following problem is from Ch. 15 of Spivak's Calculus. My question is about understanding the solution manual solution to item $(c)$. I've asked a separate question about my own attempt at a solution which differs from the solution manual.
*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.
Item $(a)$ is just the limit of a definite integral
$$\lim\limits_{\lambda\to \infty} \int_a^b \sin{(\lambda x)}dx$$
$$=\lim\limits_{\lambda\to \infty} -\frac{1}{\lambda}\left ( \cos{(\lambda d)}-\cos{(\lambda c)} \right )=0$$
For item $(b)$, note that a step function is defined based on a partition $P=\{t_0,...,t_n\}$ of $[a,b]$. Then
$$\lim\limits_{\lambda\to \infty}\int_a^b s(x)\sin{(\lambda x)}dx=\lim\limits_{\lambda\to \infty} \sum_{i=1}^n s_i\int_{t_{i-1}}^{t_i} \sin{(\lambda x)}dx=0$$
For the items above, my solution coincided with the solution manual. For item $(c)$ my solution was different. For now I'd like to understand the solution manual solution (specifically the last part of it). Here it is
Solution Manual Solution to $(c)$
For any $\epsilon>0$ there is, by Problem 13-16, a step function $s\leq f$ with
$$\int_a^b [f(x)-s(x)]dx<\epsilon$$
Now
$$\left | \int_a^b f(x)\sin{(\lambda x)}dx-\int_a^b s(x)\sin{(\lambda x)}dx \right |\tag{1}$$ $$=\left | \int_a^b [f(x)-s(x)]\sin{(\lambda x)} dx \right |$$
$$\leq \int_a^b [f(x)-s(x)]\cdot |\sin{(\lambda x)} |dx$$
$$\leq \int_a^b [f(x)-s(x)]dx<\epsilon\tag{2}$$
So far so good
Part $(b)$ then shows that
$$\lim\limits_{\lambda\to \infty} \left | \int_a^b f(x)\sin{(\lambda x)}dx \right |<\epsilon$$
Since this is true for every $\epsilon>0$, the limit must be $0$.
I'd like to understand this last part.
I think what happened is that from $(1)$ we had
$$\left | \int_a^b f(x)\sin{(\lambda x)}dx\right | -\left | \int_a^b s(x)\sin{(\lambda x)}dx \right |\leq\left | \int_a^b f(x)\sin{(\lambda x)}dx-\int_a^b s(x)\sin{(\lambda x)}dx \right |<\epsilon$$
and then we took the limit
$$\lim\limits_{\lambda\to \infty}\left | \int_a^b f(x)\sin{(\lambda x)}dx\right | -\lim\limits_{\lambda\to \infty}\left | \int_a^b s(x)\sin{(\lambda x)}dx \right |<\epsilon$$
But since from $(b)$ we know that $\lim\limits_{\lambda\to \infty}\left | \int_a^b s(x)\sin{(\lambda x)}dx\right |=0$ we have
$$\lim\limits_{\lambda\to \infty}\left | \int_a^b f(x)\sin{(\lambda x)}dx\right |<\epsilon$$
Is this filling in of the steps correct?
