Let $X_1, X_2, \cdots$ be independent identically distributed random variables having a common distribution function $F$ and let $f_n(x) = (na_n)^{-1} \sum^n_{i = 1} k((x - X_i)/a_n)$ where $\{a_n\}$ is a sequence of positive numbers converging to zero and $k$ is a probability density function. If $\sum^\infty_{n = 1} \exp(- cna_n^2)$ is finite for all positive $c$ and if $k$ satisfies: (i) $k$ is continuous and of bounded variation on $(-\infty, \infty)$. (ii) $uk(u) \rightarrow 0$ as $u \rightarrow + \infty$ or $-\infty$. (iii) There exists a $\delta$ in (0, 1) such that $u(V^{-u^\delta}_{-\infty} (k) + V^\infty_{u^\delta} (k)) \rightarrow 0$ as $u \rightarrow \infty$. (iv) $\int|u| dk(u)$, the integral of $|u|$ with respect to the signed measure determined by $k$, is finite. Then the author [2] has established the following: THEOREM. A necessary and sufficient condition for $\lim_{n\rightarrow\infty} \sup_x|f_n(x) - g(x)| = 0$ with probability one for a function $g$ is that $g$ be the uniformly continuous derivative of $F$. The purpose of this note is to show that this theorem remains true if conditions (i)-(iv) on $k$ are replaced by the condition that $k$ is of bounded variation on $(-\infty, \infty)$.