$\newcommand\ep\epsilon\newcommand\de\delta\newcommand\R{\mathbb R}$Consider any family of nonnegative measurable kernels $K_\ep\colon(0,\infty)\to\mathbb R$ for $\ep\in(0,\infty)$ such that $\int_0^a K_\ep(z)\,dz\to1$ for each real $a>0$ (and hence $\int_\de^a K_\ep(z)\,dz\to0$ each real $\de\in(0,a)$) as $\ep\downarrow0$. Then for any real $x$ and any (say locally bounded) measurable function $f\colon\R\to\R$ that is right-continuous at $x$ and any real $b>x$, we have $$\int_x^b K_\ep(y-x)f(y)\,dy\to f(x)$$ as $\ep\downarrow0$.
This follows easily by writing $\int_x^b=\int_x^{x+\de}+\int_{x+\de}^b$ for small $\de>0$.
In your case, just take $K_\ep(z)=\ep z^{\ep-1}$ for real $z>0$.
Another example of an appropriate family of kernels -- converging to the delta function as $\ep\downarrow0$ -- would be $K_\ep(z)=e^{-z/\ep}/\ep$ for real $z>0$. More generally, one can take $K_\ep(z)=k(z/\ep)/\ep$ for real $z>0$, where $k\colon(0,\infty)\to\mathbb R$ is any nonnegative measurable function such that $\int_0^\infty k(t)\,dt=1$.