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Iosif Pinelis
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$\newcommand\ep\epsilon\newcommand\de\delta\newcommand\R{\mathbb R}$Consider any family of nonnegative measurable kernels $K_\ep\colon(0,\infty)\to\mathbb R$ with $\ep\in(0,\infty)$ such that $\int_0^\infty K_\ep(z)\,dz=1$ for all $\ep$ and $\int_\de^\infty K_\ep(z)\,dz\to0$ for each real $\de>0$ 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, take $K_\ep(z)=\ep z^{\ep-1}$ for real $z>0$.

Iosif Pinelis
  • 127.7k
  • 8
  • 107
  • 229