Floor of a random variable with bounded density

Question

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.

Answer 1

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 $$\mathbb P(\lfloor \lambda X\rfloor \mbox{ is even}) = \int_{\mathbb R^+}\mathbb 1_{\{\lfloor \lambda s\rfloor\mbox{ is even}\}} f_X(s)\,\mathrm ds = \int_{\mathbb R^+} f_X(s)\, \mathrm d\mu_{\lambda}(s) \xrightarrow[\lambda \to \infty]{} \frac{1}{2}.$$