Convergence and different properties (measurable, differentiable)

Question

Let $\lambda$ be the Lebesgue measure on $\mathbb R$ and $h \in C^{\infty} (\mathbb R)$ with compact provider. Let $f_n(x) = n^{\alpha} h (nx) $

For which values of $\alpha > 0$ can we say that $f_n \rightarrow 0\ \ \lambda$-almost everywhere on $\mathbb R$?

Answer 1

I think that compact provider means compact support.

Let $K$ be a compact subset of $\mathbb R$ such that $h(x)=0$ for all $x \notin K.$

Take $x \in \mathbb R$ with $x \ne 0$. Then there ist $N$ such that $nx \notin K$ for all $n>N.$ Hence $f_n(x)=0$ for all $n>N.$

Conclusion: $f_n(x) \to 0$ as $n \to \infty$ for all $x \ne 0.$

Answer 2

If by compact provider, you mean compact support, then surely, this holds for any value of $\alpha$. Let $x\in \mathbb{R}\setminus \{0\}.$ Let $K=\sup \textrm{supp}(h)$. Then $f_n(x)=0$ for every $n\geq \frac{K}{|x|}$. Hence, $f_n(x)\to 0$. As $\{0\}$ is a Lesbegue null-set, we're done.