The Linearity Property of the Lebesgue Integral of Simple Functions

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}