# Floor of a random variable with bounded density

Let $$X$$ be a random variable on $$\mathbb R^+$$ with bounded density, and let $$\lfloor \cdot \rfloor$$ denote the floor function. Show that for $$\lambda\in \mathbb R^+$$, $$\lim_{\lambda\to\infty} \mathbb P(\lfloor \lambda X\rfloor \mbox{ is even}) = \frac{1}2.$$

The statement makes intuitive sense to me, but I don't know how to show it. I created this simple example to better understand a problem I'm working on where space is partitioned into equivalence classes, (odd / even, in the case of the example) "uniformly" spread out over the space, and fine with respect to the extent of the bounded density. If alternative assumptions on $$X$$ can be used, I would be curious to hear suggestions.

I believe I have the answer. For $$\lambda\in \mathbb R^+$$, we can define a measure on $$\mathbb R^+$$ by $$\mu_{\lambda}(A) = \int_{A}\mathbb 1_{\{\lfloor \lambda s\rfloor\mbox{ is even}\}}\,\mathrm ds,\qquad A\in\mathcal B(\mathbb R^+).$$ Remark that for $$b > a > 0$$, $$\mu_\lambda((a,b)) \xrightarrow[\lambda \to \infty]{} \frac{b-a}{2}.$$ Since $$\mu_\lambda$$ converges to half of the Lebesgue measure on the open intervals in $$\mathbb R^+$$, which generate $$\mathcal B(\mathbb R^+)$$, we have that $$\mu_\lambda$$ converges to half of the Lebesgue measure pointwise on $$\mathcal B(\mathbb R^+)$$. Moreover, this implies that $$\int_{\mathbb R^+} f(s)\,\mathrm{d}\mu_\lambda(s) \xrightarrow[\lambda\to\infty]{} \frac{1}{2}\int_{\mathbb R^+}f(s)\,\mathrm{d} s,$$ for any bounded measurable function $$f$$ (see this post). Finally, with $$f_X$$ the density of $$X$$, one has $$\mathbb P(\lfloor \lambda X\rfloor \mbox{ is even}) = \int_{\mathbb R^+}\mathbb 1_{\{\lfloor \lambda s\rfloor\mbox{ is even}\}} f_X(s)\,\mathrm ds = \int_{\mathbb R^+} f_X(s)\, \mathrm d\mu_{\lambda}(s) \xrightarrow[\lambda \to \infty]{} \frac{1}{2}.$$
• You are on the right track. Check first that, for $f \in C^{1}_c(\mathbb{R}^+)$, integration by parts yields: $$\int_{\mathbb{R}^+} f(s)\,\mu_\lambda(\mathrm{d}s)=-\int_{\mathbb{R}^+}f'(s)\mu_\lambda((0,s))\,\mathrm{d}s\to-\int_{\mathbb{R}^+}f'(s)\frac{s}{2}\,\mathrm{d}s=\frac{1}{2}\int_{\mathbb{R}^+}f(s)\,\mathrm{d}s.$$ Then, for general $f$, note that the linear functional $$L^1(\mathbb{R})\ni f\mapsto\int_{\mathbb{R}^+} f(s)\,\mu_\lambda(\mathrm{d}s)$$ is uniformly bounded, hence the above convergence extends to all of $L^1(\mathbb{R}^+)$ by density argument. Commented Mar 11 at 17:29