Lebesgue Integrability of Functions Equalling 0 a.e. on General Intervals

# Lebesgue Integrability of Functions Equalling 0 a.e. on General Intervals

If the values of the function $f$ equal $0$ almost everywhere on the interval $I$ that $f$ is defined on then it is desirable to have $f$ be Lebesgue integrable and for the Lebesgue integral of $f$ to equal $0$. Fortunately, this is indeed the case as we see in the following theorem.

 Theorem 1: Let $f$ be a function defined on $I$. If $f(x) = 0$ almost everywhere on $I$ then $f$ is Lebesgue integrable on $I$ and $\displaystyle{\int_I f(x) \: dx = 0}$.
• Proof: Let $f$ be a function defined on $I$ such that $f(x) = 0$ almost everywhere on $I$. Define the sequence $(f_n(x))_{n=1}^{\infty}$ such that for all $n \in \mathbb{N}$:
(1)
\begin{align} \quad f_n(x) = 0 \end{align}
• Then $(f_n(x))_{n=1}^{\infty}$ is an increasing sequence of functions that converges to $f$ almost everywhere on $I$. Furthermore, since $\int_I f_n(x) \: dx = 0$ for each $n \in \mathbb{N}$ we see that $\displaystyle{\lim_{n \to \infty} \int_I f_n(x) \: dx}$ is finite. So $f$ is an upper function. But every upper function on $I$ is Lebesgue integrable on $I$. $\blacksquare$
 Theorem 2: Let $f$ and $g$ be functions defined on $I$. If $f$ is Lebesgue integrable on $I$ and $f(x) = g(x)$ almost everywhere on $I$ then $g$ is Lebesgue integrable on $I$ and $\displaystyle{\int_I f(x) \: dx = \int_I g(x) \: dx}$.
• Proof: Since $f(x) = g(x)$ almost everywhere on $I$, we see that the difference $f(x) - g(x) = 0$ almost everywhere on $I$. So by Theorem 1 we have that $f - g$ is Lebesgue integrable on $I$ and furthermore:
(2)
\begin{align} \quad \int_I [f(x) - g(x)] = 0 \end{align}
• By the Linearity of Lebesgue Integrals, we have that since $f$ and $f - g$ are Lebesgue integrable on $I$ that then $f - (f - g) = g$ is Lebesgue integrable on $I$ and additionally:
(3)
\begin{align} \quad \int_I f(x) \: dx = \int_I f(x) \: dx - \underbrace{\int_I [f(x) - g(x)] \: dx}_{=0} = \int_I g(x) \: dx \end{align}
• So $\displaystyle{\int_I f(x) \: dx = \int_I g(x) \: dx}$ as desired. $\blacksquare$