Let $f\in L^1(\mathbb{R^n})$ with $\int_{\mathbb{R}^n}fdx=1$ and let $f_{\alpha}(x)=\alpha^n f(\alpha x)$ for $\alpha>0$. Suppose that $\phi:\mathbb{R}^n\to\mathbb{R}$ is continuous in $0$. Is it true that without the assumption of boundedness or integrability on $\phi$ we still have $\lim_{\alpha\to \infty}\int_{\mathbb{R^n}}f_\alpha(x)\phi(x)dx=\phi(0)$?
For $\phi\in C_c^\infty(\mathbb{R}^n)$ the result can be proven with dominated convergence.
Another approach would be to choose $\delta>0$ for $\epsilon >0$ such that $|\phi(x)|<\epsilon$ on $B_\delta(0)$, then $$\left\lvert\int f_\alpha\phi dx-\phi(0) \right\rvert= \left\lvert \int f_\alpha(x) (\phi(x)-\phi(0))dx \right\rvert \le \epsilon+\int_{B_{\alpha\delta}(0)^c} |f(x)||\phi(\tfrac{x}{\alpha})-\phi(0)|dx. $$ Surely the measure theoretic limit of $B_{\alpha\delta}(0)^c$ is $\emptyset$ but does this yield that the integral goes to zero too without knowing that $\phi$ is integrable?