Linearity of Lebesgue Integrals

# Linearity of Lebesgue Integrals

Recall from The Lebesgue Integral page that a function $f$ is said to be Lebesgue integrable on the interval $I$ if $f$ is equal to the difference of two upper functions on $I$, that is, there exists $u, v \in U(I)$ such that $f = u - v$.

Furthermore, we define the Lebesgue integral of $f$ on $I$ to be:

(1)\begin{align} \quad \int_I f(x) \: dx = \int_I u(x) \: dx - \int_I v(x) \: dx \end{align}

We noted that the value of $\int_I f(x) \: dx$ is independent of the choices of upper functions $u$ and $v$ on $I$ such that $f = u - v$.

We will now show that the linearity properties hold for Lebesgue integrals.

Theorem 1: Let $f$ and $g$ be Lebesgue integrable on $I$. Then $f + g$ is Lebesgue integrable on $I$ and $\displaystyle{\int_I [f(x) + g(x)] \: dx = \int_I f(x) \: dx + \int_I g(x) \: dx}$. |

**Proof:**Let $f$ and $g$ be Lebesgue integrable on $I$. Then there exists upper functions $u_1, u_2, v_1, v_2$ on $I$ such that:

\begin{align} \quad f = u_1 - v_1 \quad g = u_2 - v_2 \end{align}

- So $f + g = (u_1 + u_2) - (v_1 + v_2)$. We know that $u_1 + u_2$ and $v_1 + v_2$ are upper functions, so $f + g$ is Lebesgue integrable on $I$ and furthermore:

\begin{align} \quad \int_I [f(x) + g(x)] \: dx &= \int_I [u_1(x) + u_2(x)] \: dx - \int_I [v_1(x) + v_2(x)] \: dx \\ \quad &= \int_I u_1(x) \: dx + \int_I u_2(x) \: dx - \int_I v_1(x) \: dx - \int_I v_2(x) \: dx \\ \quad &= \left ( \int_I u_1(x) \: dx - \int_I v_1(x) \: dx \right ) + \left ( \int_I u_2(x) \: dx - \int_I v_2(x) \: dx \right ) \\ \quad & = \int_I f(x) \: dx + \int_I g(x) \: dx \end{align}

Theorem 2: Let $f$ be Lebesgue integrable on $I$. Then for all $c \in \mathbb{R}$, $cf$ is Lebesgue integrable on $I$ and $\int_I cf(x) \: dx = c \int_I f(x) \: dx$. |

**Proof:**Let $f$ be Lebesgue integrable on $I$. Then there exists upper functions $u, v$ on $I$ such that $f = u - v$. Then $cf = cu - cv$. If $c \geq 0$ then $cu$ and $cv$ are still upper functions. If $c < 0$, then $cu - cv = (-cv) - (-cu)$ and so $-c > 0$ and so in both cases we see that $cf$ is Lebesgue integrable on $I$. So:

\begin{align} \quad \int_I cf(x) \: dx &= \int_I cu(x) \: dx - \int_I cv(x) \: dx \\ \quad &= c \left ( \int_I u(x) \: dx - \int_I v(x) \: dx \right ) \\ \quad &= c \int_I f(x) \: dx \quad \blacksquare \end{align}