Functions of Lebesgue Integrals

Let $f$ be a Lebesgue integrable function on $[a, b]$. We will define a new function $F : [a, b] \to \mathbb{R}$ for all $x \in [a, b]$.

(1)

\begin{align} \quad F(x) = \int_a^x f(t) \: dt \end{align}

Note that $\displaystyle{F(A) = \int_a^a f(t) \: dt = \int_{\{a\}} f(t) \: dt = 0}$.

We now state two important properties of $F$.

Theorem 1: If $f$ is Lebesgue integrable on $[a, b]$ and $\displaystyle{F(x) = \int_a^x f(t) \: dt}$ then $F$ is uniformly continuous on $[a, b]$ and hence continuous on $[a, b]$.

• Proof: Let $\epsilon > 0$ be given. Then there exists a $\delta > 0$ such that if $E$ is a Lebesgue measurable set in $[a, b]$ with $m(E) < \delta$ then:

(2)

\begin{align} \quad \int_E |f(x)| < \epsilon \end{align}

• So for $\epsilon > 0$ if $x_1, x_2 \in [a, b]$ are such that $|x_2 - x_1| < \delta$ (and assuming without loss of generality that $x_1 < x_2$ then:

(3)

\begin{align} \quad | F(x_2) - F(x_1) | &= \biggr \lvert \int_a^{x_2} f(t) \: dt - \int_a^{x_1} f(t) \: dt \biggr \rvert \\ & \leq \biggr \lvert \int_{x_1}^{x_2} f(t) \: dt \biggr \rvert \\ & \leq \biggr \lvert \int_{[x_1, x_2]} f(t) \: dt \biggr \rvert \\ & \leq \int_{[x_1, x_2]} | f(t) | \: dt \\ & < \epsilon \end{align}

• So for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $x_1, x_2 \in [a, b]$ and $| x_2 - x_1 | < \delta$ then $| F(x_2) - F(x_1) | < \epsilon$, so $F$ is uniformly continuous on $[a, b]$ and hence $F$ is continuous on $[a, b]$. $\blacksquare$

Theorem 2: If $f$ is Lebesgue integrable on $[a, b]$ and $\displaystyle{F(x) = \int_a^x f(t) \: dt}$ then $F$ is of bounded variation on $[a, b]$.

• Proof: Let $P = \{ a = x_0, x_1, ..., x_n = b \}$ be a partition on $[a, b]$ and let $\displaystyle{M = \int_a^b | f(t) | \: dt}$. Since $f$ is Lebesgue integrable on $[a, b]$ we have that $M < \infty$ by definition. Now consider the variation of $f$ associated with $P$:

(4)

\begin{align} \quad V(P, F) &= \sum_{i=1}^{n} | F(x_i) - F(x_{i-1}) | \\ &= \sum_{i=1}^{n} \biggr \lvert \int_a^{x_i} f(t) \: dt - \int_a^{x_{i-1}} f(t) \: dt \biggr \rvert \\ &= \sum_{i=1}^{n} \biggr \lvert \int_{x_{i-1}}^{x_i} f(t) \: dt \biggr \rvert \\ & \leq \sum_{i=1}^{n} \int_{x_{i-1}}^{x_i} |f(t)| \: dt \\ & \leq \int_a^b | f(t) | \: dt \\ & \leq M \end{align}

• So for all partitions $P \in \mathscr{P}[a, b]$ we have that $V(P, F) \leq M$ and so $F$ is of bounded variation on $[a, b]$. $\blacksquare$