If $f:\mathbb{R} \rightarrow \overline{\mathbb{R}}$ is integrable, I want to show $\sum_{n \in \mathbb{N}} f(n^ax+cos(n))$ converges for a.e. $x \in \mathbb{R}$ and $a>1$
I want to show $\lim \limits_{t \rightarrow 1} \int |g(x)(f(x)-f(tx))| dx = 0$ given $f \in L_1(\mathbb{R})$ and $f \in L_1(\mathbb{R})$ and $g \in L_{\infty}(\mathbb{R}) $.
Attempt: I can show that $\lim \limits_{t \rightarrow 1} \int |g(x)(f(x)-f(tx))| dx \le ||g||_{\infty}\lim \limits_{t \rightarrow 1} \int |(f(x)-f(tx))| dx$ and if I simply evaluate the limit, I get $||g||_{\infty} \int|f(x)-f(x))| dx = 0$. Is this enough?
For 1., use the transformation $t=n^ax+\cos n$ with the integral $\int |f(n^ax+\cos n)| dx = \frac{1}{n^a} \int |f|$. Tonelli gives $\int \sum_{n \in \mathbb{N}} |f(n^ax+\cos n)| dx =\sum_{n \in \mathbb{N}} \int |f(n^ax+\cos n)| dx = \sum_{n \in \mathbb{N}} \frac{1}{n^a} \int |f| < \infty$. It follows that $ x \mapsto \sum_{n \in \mathbb{N}} |f(n^ax+\cos n)|$ is finite a.e., from which the desired result follows.
For 2., you already have observed that $\int |g(x)(f(x)-f(tx))| dx \le \|g\|_\infty \int |f(x)-f(tx)| dx$. Let $f_t(x) = f(tx)$, then to finish, you need to show that the map $ t \mapsto f_t$ is continuous at $t=1$.
As an aside, note that $\int f_t = \frac{1}{t} \int f$.
If $g \in C_c(\mathbb{R})$ (continuous functions with compact support), then a uniform continuity argument shows that $\lim_{t \to 1}\|g - g_t\|_1 = 0$. Since $C_c(\mathbb{R})$ is dense in $L^1(\mathbb{R})$, if $ \epsilon >0$ we can find $g \in C_c(\mathbb{R})$ such that $\|f-g\|_1 < \frac{\epsilon}{3}$. Then we have the estimate \begin{eqnarray} \|f-f_t\|_1 &\le& \|f-g\|_1 + \|g-g_t\|_1 + \|f_t-g_t\|_1\\ &=& (1+\frac{1}{t}) \|f-g\|_1 + \|g-g_t\|_1 \\ &\le& (1+\frac{1}{t}) \frac{\epsilon}{3} + \|g-g_t\|_1 \end{eqnarray} Consequently, we can find a $\delta>0$ such that if $|1-t| < \delta$, then $\|f-f_t\|_1 < \epsilon$. Hence $ t \mapsto f_t$ is continuous at $t=1$.