Comparison Theorems for Lebesgue Integrals

# Comparison Theorems for Lebesgue Integrals

Recall from the The Lebesgue Integral page that a function $f$ on $I$ is said to be Lebesgue integrable on $I$ if $f$ is the difference of two upper functions, that is, there exists upper functions $u$ and $v$ on $I$ such that $f = u - v$. We denoted the set of all Lebesgue integrable functions on $I$ by $L(I)$. Furthermore, we defined 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 will now look at some nice comparison theorems for Lebesgue integrals.

Theorem 1: Let $f$ be Lebesgue integrable on $I$. Then if $f(x) \geq 0$ almost everywhere on $I$ then $\displaystyle{\int_I f(x) \: dx \geq 0}$. |

**Proof:**Let $f$ be Lebesgue integrable on $I$. Then there exists upper functions $u$ and $v$ on $I$ such that $f = u - v$. Since $f(x) \geq 0$ almost everywhere on $I$, this implies that $u(x) - v(x) \geq 0$ almost everywhere on $I$, i.e., $u(x) \geq v(x)$ almost everywhere on $I$.

- So, by one of theorems presented on the A Comparison Theorem for Integrals of Upper Functions on General Intervals page, we have that:

\begin{align} \quad \int_I u(x) \: dx \geq \int_I v(x) \: dx \end{align}

- Thus we have that:

\begin{align} \quad \int_I f(x) \: dx = \int_I u(x) \: dx - \int_I v(x) \: dx \geq 0 \quad \blacksquare \end{align}

Theorem 2: Let $f$ and $g$ be Lebesgue integrable on $I$. If $f(x) \leq g(x)$ almost everywhere on $I$ then $\displaystyle{\int_I f(x) \: dx \leq \int_I g(x) \: dx}$. |

**Proof:**Let $f$ and $g$ be Lebesgue integrable on $I$. Consider the function $h = g - f$. This function is Lebesgue integrable from the Linearity of Lebesgue Integrals since $h = g + (-1)f$. Furthermore, since $f(x) \leq g(x)$ almost everywhere on $I$ we have that $h(x) = g(x) - f(x) \geq 0$ almost everywhere on $I$. Applying Theorem 1 and we have that:

\begin{align} \quad 0 \leq \int_I h(x) \: dx = \int_I [g(x) - f(x)] \: dx = \int_I f(x) \: dx - \int_I g(x) \: dx \end{align}

- Therefore $\displaystyle{\int_I f(x) \: dx \leq \int_I g(x) \: dx}$. $\blacksquare$

Theorem 3: Let $f$ and $g$ be Lebesgue integrable on $I$. If $f(x) = g(x)$ almost everywhere on $I$ then $\displaystyle{\int_I f(x) \: dx = \int_I g(x) \: dx}$. |

**Proof:**Let $f$ and $g$ be Lebesgue integrable on $I$ and suppose that $f(x) = g(x)$ almost everywhere on $I$. Then $f(x) \leq g(x)$ almost everywhere on $I$ and by applying Theorem 2 we see that:

\begin{align} \quad \int_I f(x) \: dx \leq \int_I g(x) \: dx \quad (*) \end{align}

- Furthermore, $f(x) \geq g(x)$ almost everywhere on $I$ and by applying THeorem 2 again we see that:

\begin{align} \quad \int_I f(x) \: dx \geq \int_I g(x) \: dx \quad (**) \end{align}

- Combining $(*)$ and $(**)$ shows us that $\displaystyle{\int_I f(x) \: dx = \int_I g(x) \: dx}$. $\blacksquare$