Calculate
$\lim_{n \rightarrow \infty} \int_{[0, \infty]} e^{-\sqrt x}$ $\cos$ ($x^2 \over n$) $d\lambda^1(x)$.
My attempt:
First, we want to use the Dominated Convergence Theorem. We define
$f_n(x) := e^{-\sqrt x}$ $\cos$($x^2 \over n$)
and note that this sequence of functions is integrable.
Furthermore, we define $f := \lim_{n \rightarrow \infty} e^{-\sqrt x}$ $\cos$ ($x^2 \over n$) $= e^{-\sqrt x}$.
Since $|e^{-\sqrt x}$ $\cos$($x^2 \over n$)| $\le e^{-\sqrt x}$, we set
$F := f$, which shall be our majorant.
Now we have everything that we need in order to apply the Dominated Convergence Theorem. Thus, the above integral is identical to
$\int_{[0, \infty]} e^{-\sqrt x} d\lambda(x)$.
Since $[0, \infty] = \lim_{m \rightarrow \infty} [0, m]$, we are allowed to write this as
$\lim_{m \rightarrow \infty} \int_{[0, m]} e^{-\sqrt x} d\lambda(x)$ and treat is as a Riemann-integral:
$\lim_{m \rightarrow \infty} \int_0^m e^{-\sqrt x} dx$.
Integrating delivers
$[-2(\sqrt x + 1) e^{-\sqrt x}]_0^m = -2(\sqrt m + 1) e^{-\sqrt m} + 2$, and by taking the limit $m \rightarrow \infty$, the expression $-2(\sqrt m + 1) e^{-\sqrt m}$ converges to $0$, thus,
$\lim_{m \rightarrow \infty} \int_0^m e^{-\sqrt x} dx = 2.$
