Help understanding approximation of integral of pdf

Question

Assume that $f$ is the pdf of a continuous random variable $X:\Omega\to\mathbb{R}$. Let $\varepsilon>0$. Then: \begin{equation*} \mathbb{P}\left(X\in\left[x-\frac{\varepsilon}{2},x+\frac{\varepsilon}{2}\right]\right)=\int_{x-\frac{\varepsilon}{2}}^{x+\frac{\varepsilon}{2}}f(t)\,\mathrm{d}t\approx \varepsilon f(x), \quad x\in\mathbb{R}. \end{equation*}

My question: How does one realize that the approximation above is true for sufficiently small $\varepsilon$?

Answer 1

Assume $F(t)=\int f(t)dt$. Therefore

$$\int_{x-\frac{\varepsilon}{2}}^{x+\frac{\varepsilon}{2}}f(t)\,\mathrm{d}t = F\left(x+\frac{\varepsilon}{2}\right)-F\left(x-\frac{\varepsilon}{2}\right)$$

Now by expanding $F$ as $F\left(x+\frac{\varepsilon}{2}\right)\approx F(x)+f(x)\frac{\varepsilon}{2} + O(\varepsilon^2)$ and $F\left(x-\frac{\varepsilon}{2}\right)\approx F(x)-f(x)\frac{\varepsilon}{2} + O(\varepsilon^2)$ and using the first order expansion only, we get

$$F\left(x+\frac{\varepsilon}{2}\right)-F\left(x-\frac{\varepsilon}{2}\right)\approx F(x)+f(x)\frac{\varepsilon}{2} - F(x)+f(x)\frac{\varepsilon}{2} = f(x)\varepsilon$$