The Linearity Property of the Lebesgue Integral of Simple Functions

Recall from The Lebesgue Integral of Simple Functions page that if $\varphi$ is a simple function defined on a Lebesgue measurable set $E$ with $m(E) < \infty$ with canonical representation $\displaystyle{\varphi(x) = \sum_{k=1}^{n} a_k \chi_{E_k}(x)}$ then the Lebesgue integral of $\varphi$ on $E$ is defined as:

(1)

\begin{align} \quad (L) \int_E \varphi = \sum_{k=1}^{n} a_k m(E_k) \end{align}

We also proved that if $E_1, E_2, ..., E_n$ are disjoint subsets of $E$ with $\displaystyle{\varphi_(x) = \sum_{k=1}^{n} a_i \chi_{E_k}(x)}$ then also $\displaystyle{(L) \int_E \varphi = \sum_{k=1}^{n} a_k m(E_k)}$.

We will now show that the Lebesgue integral of simple functions has the linearity property.

Theorem 1 (Linearity of the Lebesgue Integral of Simple Functions): Let $\varphi$ and $\psi$ be simple functions defined on a Lebesgue measurable set $E$ with $m(E) < \infty$. Then for all $\alpha, \beta \in \mathbb{R}$ we have that $\displaystyle{\int_E (\alpha \varphi + \beta \psi) = \alpha \int_E \varphi + \beta \int_E \psi}$.

We make use of the theorem mentioned above to show that the Lebesgue integral of simple functions has the linearity property.

• Proof: Let $E_1, E_2, ..., E_n$ be disjoint subsets of $E$ such that $\displaystyle{\varphi(x) = \sum_{k=1}^{n} a_k \chi_{E_k}(x)}$ and $\displaystyle{\psi(x) = \sum_{k=1}^{n} b_k \chi_{E_k}(x)}$. Then for any $\alpha, \beta \in \mathbb{R}$ we have that:

(2)

\begin{align} \quad \alpha \varphi(x) + \beta \psi (x) = \sum_{k=1}^{n} (\alpha a_k + \beta b_k) \chi_{E_k} \end{align}

• Therefore:

(3)

\begin{align} \quad \int_E (\alpha \varphi + \beta \psi) &= \sum_{k=1}^{n} (\alpha a_k + \beta b_k) m(E_k) \\ &= \sum_{k=1}^{n} \alpha a_k m(E_k) + \sum_{k=1}^{n} \beta b_k m(E_k) \\ &= \alpha \sum_{k=1}^{n} a_k m(E_k) + \beta \sum_{k=1}^{n}b_k m(E_k) \\ &= \alpha \int_E \varphi + \beta \int_E \psi \quad \blacksquare \end{align}