Lebesgue Integrable Functions with Arbitrarily Small Integral Terms

# Lebesgue Integrable Functions with Arbitrarily Small Integral Terms

We will soon prove a very important result called the Riemann-Lebesgue lemma but we will first need to look at a result regarding Lebesgue integrable functions. If $f \in L(I)$, then as the following theorem will prove, for any $\epsilon > 0$ there will exist upper functions $u^*, v^* \in S(I)$ such that $f = u^* - v^*$, $v^*$ is nonnegative almost everywhere on $I$, and the integral of $v^*$ on $I$ is less than $\epsilon$.

Similarly, for all $\epsilon > 0$ there exists functions $g \in L(I)$, $s \in S(I)$ such that $f = g + s$ and where the integral of the absolute value of $g$ on $I$ is less than $\epsilon$.

Lemma 1: Let $f \in L(I)$. Then:a) For all $\epsilon > 0$ there exists upper functions $u^*, v^* \in U(I)$ such that $f = u^* - v^*$, $v^*$ is nonnegative almost everywhere on $I$, and $\int_I v^*(x) \: dx < \epsilon$.b) For all $\epsilon > 0$ there exists functions $g \in L(I)$ and $s \in S(I)$ such that $f = g + s$ and $\int_I \mid g(x) \mid \: dx < \epsilon$. |

**Proof of a)**Let $\epsilon > 0$ be given. Since $f \in L(I)$ there exists upper functions $u, v \in U(I)$ such that:

\begin{align} \quad f = u - v \end{align}

- Since $v$ is an upper function there exists an increasing sequence of step functions $(v_n(x))_{n=1}^{\infty}$ that converges to $v$ almost everywhere on $I$ and such that $\displaystyle{\lim_{n \to \infty} \int_I v_n(x) \: dx}$ is finite. Moreover, since $(v_n(x))_{n=1}^{\infty}$ converges to $v$ almost everywhere on $I$ we have that for this given $\epsilon > 0$ that there exists an $N^* \in \mathbb{N}$ such that if $n \geq N^*$ then:

\begin{align} \quad \mid v(x) - v_n(x) \mid < \epsilon \end{align}

- If $l > 0$ denotes the length of the interval $I$, then there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $\mid v(x) - v_n(x) \mid < \frac{\epsilon}{l}$ and so:

\begin{align} \quad \int_I \mid v(x) - v_N(x) \mid \: dx < \int_I \frac{\epsilon}{l} \: dx = l \cdot \frac{\epsilon}{l} = \epsilon \end{align}

- So let $u^* = u - v_N$ and $v^* = v - v_N$. Then $u^*, v^* \in U(I)$, $f = (u - v_N) - (v - v_N) = u^* - v^*$, $v$ is nonnegative almost everywhere on $I$ (since $(v_n(x))_{n=1}^{\infty}$ is an increasing sequence that converges to $v(x)$ almost everywhere on $I$), and moreover, $\int_I v^*(x) < \epsilon$. $\blacksquare$

**Proof of b)**Let $\epsilon > 0$ be given. Since $f in L(I)$, by part (a), for $\displaystyle{\epsilon_1 = \frac{\epsilon}{2} > 0}$ there exists upper functions $u^*$ and $v^*$ with $f = u^* - v^*$, $v^*$ nonnegative almost everywhere on $I$, and:

\begin{align} \quad \int_I v^*(x) \: dx = \int_I v^*(x) \: dx < \epsilon_1 = \frac{\epsilon}{2} \quad (*) \end{align}

- For the function $u^*$ choose $s \in S(I)$ such that for $\displaystyle{\epsilon_2 = \frac{\epsilon}{2} > 0}$ we have that:

\begin{align} \quad \int_I \mid u^*(x) - s(x) \mid < \epsilon_2 = \frac{\epsilon}{2} \quad (**) \end{align}

- Then let $g = u^* - s - v^*$. Then $f = (u^* - s - v^*) + s = g + s$, $g \in L(I)$, and:

\begin{align} \quad \quad \int_I \mid g(x) \mid \: dx = \int_I \mid u^*(x) - s(x) - v^*(x) \mid \: dx \leq \int_I [\mid u^*(x) - s(x) \mid + \mid v^*(x) \mid] \: dx \leq \int_I \mid u^*(x) - s(x) \mid \: dx + \int_I v^*(x) \: dx < \epsilon_1 + \epsilon_2 = \epsilon \end{align}

- So the function $g$ defined as above suffices. $\blacksquare$