Let $(X, \mathcal{F}, \mu)$ be a measure space and $s = \sum_{i=1}^{n} a_i \chi_{A_i}$ be a non-negative simple function in canonical form (where $a_i \geq 0$ and $A_i$ are disjoint measurable sets). The Lebesgue integral of $s$ is defined as: $\int_X s \, d\mu = \sum_{i=1}^{n} a_i \mu(A_i)$

If $E \in \mathcal{F}$, we define: $\int_E s \, d\mu = \int_X s \chi_E \, d\mu = \sum_{i=1}^{n} a_i \mu(A_i \cap E)$

Let $f: X \to [0, \infty]$ be a non-negative measurable function. The Lebesgue integral of $f$ is: $\int_X f \, d\mu = \sup \left\{\int_X s \, d\mu : s \text{ is simple and } 0 \leq s \leq f\right\}$

The value may be $+\infty$. We say $f$ is integrable if $\int_X f \, d\mu < \infty$.

For a measurable function $f: X \to \mathbb{R}$, define the positive and negative parts: $f^+ = \max(f, 0), \quad f^- = \max(-f, 0)$

Then $f = f^+ - f^-$ and $|f| = f^+ + f^-$. Both $f^+$ and $f^-$ are non-negative measurable functions.

If at least one of $\int_X f^+ \, d\mu$ or $\int_X f^- \, d\mu$ is finite, we define: $\int_X f \, d\mu = \int_X f^+ \, d\mu - \int_X f^- \, d\mu$

We say $f$ is integrable if both integrals are finite, equivalently if $\int_X |f| \, d\mu < \infty$.