Show that $\displaystyle \lim_{k\rightarrow \infty}f(k^\alpha x) \rightarrow 0$ Let $f$ be an integrable function over the positive real numbers. Prove that $f(k^\alpha x) \rightarrow 0$ whenever $k \rightarrow \infty$ for almost everywhere $x>0$ and any real number $\alpha>1$.
I have two attempts:
(1) Use monotone convergence theorem for series.
Since \begin{align*}
\int \sum_{k=1}^\infty f(k^\alpha x)=\sum_{k=1}^\infty \int f(k^\alpha x) dx &= \sum_{k=1}^\infty \int \frac{1}{k^\alpha} f(y)dy \\
&=\sum_{k=1}^\infty \frac{1}{k^\alpha}  \int f(y)dy\\
&=\sum_{k=1}^\infty \frac{1}{k^\alpha}||f||_1<\infty 
\end{align*}
Then we have the integrand $\sum_{k=1}^\infty f(k^\alpha x)<\infty$ for almost everywhere $x>0$. Thus, the general term $f(k^\alpha x)$ goes to zero for almost everywhere $x>0$.
I think this attempt is valid.
(2) However I am trying a second solution:
Let $E_n=\mathbb{R}_+ \cap (-n, n)$ which has finite measure (so that I can use the continuity of measure from below and thus Borel-Cantelli lemma). Now consider the following set:
$$\{x\in E_n\mid \lim_{k\rightarrow \infty} f(k^\alpha x)=0\},$$
I want to show that for each $k\in \mathbb{N}$ and $\alpha>1$, the set
$$A_{k, \alpha}=\{x\in E_n| |f(k^\alpha x)|> M\}$$
has measure zero on each $E_n$, where $M$ is any positive real number depends on $k, \alpha$.
Choose $M=\frac{1}{k^{\alpha-2}}$ , then I want to use Markov's inequality to show that
\begin{align*}
\mu(A_{k, \alpha})=\mu\{x \in E_n| |f(k^\alpha x)|> \frac{1}{k^{\alpha-2}}\}&<k^{\alpha-2}||f||_1 \\
&= k^{\alpha-2} \int \frac{1}{k^{\alpha}} f(y) dy \\
&=\frac{1}{k^2} ||f||_1 
\end{align*}
Then by Borel-Cantelli lemma,
$$\mu(\limsup A_k)=\mu(\bigcap_{N=1}^\infty \bigcup_{k=N}^\infty A_k)=0$$
Can somebody help me to validate this approach?
 A: Following the discussion in the comments, let me try to carefully write out a solution using your approach $(2)$.
Let $\lambda > 0$ and for each positive integer $k$, define $E^{\lambda}_k = \{x \in (0, \infty): |f(k^{\alpha}x)| > \lambda\}$. By Markov's inequality,
$$
\mu(E^{\lambda}_k) \le \frac{1}{\lambda} \int_0^{\infty}|f(k^{\alpha}x)|\ \mathrm{d}\mu(x) = \frac{1}{\lambda k^{\alpha}} \int_0^{\infty}|f(x)|\ \mathrm{d}\mu(x),
$$
where the last equality follows by using the transformation $k^{\alpha}x \mapsto x$.
Since $\alpha > 1,$ we have
$$
\sum_{k=1}^{\infty}\mu(E^{\lambda}_k) \leq \sum_{k=1}^{\infty} \frac{1}{\lambda k^{\alpha}} \int_0^{\infty}|f(x)|\ \mathrm{d}\mu(x) < \infty.
$$
By the Borel-Cantelli lemma, $\mu(\{x \in (0, \infty): x \in E^{\lambda}_k \text{ for infinitely many  } k\}) = 0$, and this is true for any $\lambda > 0$.
Note that for a sequence $g_k$, we have  $g_k \not\to 0$ if and only if there exists a positive integer $n$ such that $|g_k| > \frac{1}{n}$ for infinitely many $k$. From the above discussion, we see that for every $n$, if we denote $E_n = \{x \in (0, \infty): x \in E^{1/n}_k \text{ for infinitely many  } k\}$, then  $\mu(E_n) = 0$. Since $\mu(E_n) = 0$ for each $n$, $\mu(\bigcup_{n=1}^{\infty} E_n) = 0$. Observe that
$$
\begin{align*}
\bigcup_{n=1}^{\infty} E_n &= \{x \in (0, \infty): \exists n \in \mathbb{N} \text{ such that } |f(k^{\alpha}x)| > \frac{1}{n} \text{ for infinitely many }k\}\\ &= \{x \in (0, \infty): f(k^{\alpha}x) \not\to 0\}
\end{align*}
$$
In other words, except for the set $\bigcup_{n=1}^{\infty} E_n$, which has measure zero, $f(k^{\alpha}x) \to 0$, and the conclusion follows.
I hope this is clear.
