The Lebesgue Integral

# The Lebesgue Integral

Recall that if $f$ and $g$ are both upper functions defined on the interval $I$ then the sum $f + g$ is also an upper function on $I$ and furthermore:

(1)
\begin{align} \quad \int_I [f(x) + g(x)] \: dx = \int_I f(x) \: dx + \int_I g(x) \: dx \end{align}

In general, if $f$ and $g$ are both upper functions on $I$ then the difference $f - g$ is not necessarily an upper function on $I$. Hence we define a larger class of functions and define the integral of these functions accordingly.

 Definition: A function $f$ is said to be Lebesgue Integrable on the interval $I$ if $f = u - v$ where $u$ and $v$ are both upper functions on $I$. The Set of All Lebesgue Integrable Functions on $I$ is denoted $L(I)$. Furthermore, we Lebesgue Integral of $f$ on $I$ is defined as $\displaystyle{\int_I f(x) \: dx = \int_I u(x) \: dx - \int_I v(x) \: dx}$.

Notice that from the definition above that $U(I) \subset L(I)$ for all intervals $I$. In other words, the set of all Lebesgue integrable functions on $I$ is larger than the set of all upper functions on $I$. This is very easy to see. Let $f$ be any upper function on $I$. Then $0$ is also an upper function on $I$ and $f = f - 0$, so $f$ is a Lebesgue integrable function on $I$.

Of course, we need to make one point clear. If $f$ is a Lebesgue integrable function on $I$ then $f$ may be able to be expressed as the difference of two upper functions in many different ways. Does the Lebesgue integral of $f$ on $I$ always equal the same value regardless? Fortunately the answer is yes as we prove in the following theorem.

 Theorem 1: Let $f$ be a Lebesgue integrable function such that $f = u_1 - v_1$ and $f = u_2 - v_2$ where $u_1, u_2, v_1, v_2$ are all upper functions on $I$. Then $\displaystyle{\int_I u_1(x) \: dx - \int_I v_1(x) : dx = \int_I u_2(x) \: dx - \int_I v_2(x) \: dx}$.
• Proof: Since $f = u_1 - v_1$ and $f = u_2 - v_2$ we see that:
(2)
\begin{align} \quad u_1 - v_1 = u_2 - v_2 \\ \quad u_1 + v_2 = u_2 + v_1 \end{align}
(3)
\begin{align} \quad \int_I [u_1(x) + v_2(x)] \: dx &= \int_I [u_2(x) + v_1(x)] \: dx \\ \quad \int_I u_1(x) \: dx + \int_I v_2(x) \: dx &= \int_I u_2(x) \: dx + \int_I v_1(x) \: dx \\ \quad \int_I u_1(x) \: dx - \int_I v_1(x) \: dx &= \int_I u_2(x) \: dx - \int_I v_2(x) \: dx \quad \blacksquare \end{align}