Definition. The Lebesgue integral of a measurable $f \geq 0$ against a measure $\mu$ can be defined as the increasing limit

$$\int f\,d\mu = \lim_{n \to \infty} 2^{-n}\int \lfloor 2^n f \rfloor\,d\mu$$

where the integral of any $\varphi: X \to {\mathbb N}$ is defined as

$$\int \varphi\,d\mu = \sum_{n \geq 0} n\mu(\varphi^{-1}(n)).$$

Integration of signed, complex, and vector-valued functions is defined via combinations of nonnegative real integrals.

Proposition. Let $f(x) \in o(x^k)$, that is, $\lim_{x \to 0} |f(x)/x^k| = 0$. Then

$$\int_0^x f(t)\,dt \in o(x^{k+1})$$

Proof. We have $f(x)/x^k \in o(1)$. So for $\varepsilon > 0$ let $\delta > 0$ be such that $|x| < \delta \implies |f(x)/x_k| < \varepsilon$

\begin{align*} \left|\int_0^\delta f(t)\,dt\right| &\leq \int_0^\delta |f(t)|\,dt \\ &= \int_0^\delta x_k |f(x)/x_k|\,dt \\ &\leq \int_0^\delta x_k \varepsilon\,dt \\ &= \varepsilon x^{k+1}/(k+1) < \varepsilon x^{k+1}. \end{align*}

That is, we have proved that for any $\varepsilon > 0$ there exists $\delta > 0$ such that $|\int_0^\delta f(t)\,dt| \leq \varepsilon x^{k+1}$. Which is precisely $\int_0^\delta f(t)\,dt \in o(x^{k+1})$. $\square$

Theorem. Let $f: {\mathbb R} \to {\mathbb R}$ have derivatives $f^{(k)}(0) = c_k$ for $k = 0, \dots, K$. Then:

$$f(x) = c_0 + c_1x + c_2\frac{x}2 + \dots + c_K\frac{x^K}{K!} + o(x^K)$$

Proof. The fact $f^{(K)}(0) = c_K$ implies that $f^{(K-1)}(x) = c_{K-1} + xc_K + o(x)$. Therefore, by the lemma:

\begin{align*} f^{(K-2)}(x) &= f^{(K-1)}(0) + \int_0^x f^{(K-1)}(t)\,dt \\ &= c_{K-2} + xc_{K-1} + c_Kx^2/2 + o(x^2) \end{align*}

Which can be repeated all the way down to the result. $\square$

For example, $f(x) = f(0) + f’(0)x + f’’(0)(x^2/2) + o(x^2)$.

Theorem. Additionally assume that $|f^{(K)}(x)| \leq C_K$ on the interval $x \in [0, r]$. Then:

$$x \in [0, r] \implies f(x) \leq c_0 + c_1x + \dots + C_K\frac{x_K}{K!}$$

Proof. For $x \in [0, r]$, we have:

\begin{align*} f^{(K-1)}(x) - c_{K-1} &= \int_0^x f^{(K)}(t)\,dt \\ &\leq \int_0^x C_K\,dt \\ &= xC_K. \end{align*}

hence $f^{(K-1)}(x) \leq c_{K-1} + xC_K$ and

\begin{align*} f^{(K-2)}(x) - c_{K-2} &= \int_0^x f^{(K-1)}(t)\,dt \\ &\leq \int_0^x c_{K-1} + xC_K\,dt \\ &= xc_{K-1} + (x^2/2)C_K. \end{align*}

which can be repeated all the way down to the result. $\square$

Remark: The more informative variant of this uses mean value theorem.