Integral Criteria for Functions to be Zero Almost Everywhere

# Integral Criteria for Functions to be Zero Almost Everywhere

Lemma 1: If $f$ is a nonnegative Lebesgue integrable function defined on a Lebesgue measurable set $E$ and $\displaystyle{\int_E f = 0}$ then $f(x) = 0$ almost everywhere on $E$. |

**Proof:**Let $\{ E_n \}_{n=1}^{\infty}$ be a countable collection of disjoint Lebesgue measurable sets with the property that $m(E_n) < \infty$ for each $n \in \mathbb{N}$ and $\displaystyle{E = \bigcup_{n=1}^{\infty} E_n}$. Since $f$ is a nonnegative function we have that for each $n \in \mathbb{N}$ that:

\begin{align} \quad \int_{E_n} f \leq \int_E f = 0 \end{align}

- Therefore $\displaystyle{\int_{E_n} f = 0}$ for each $n \in \mathbb{N}$. For each $n \in \mathbb{N}$ and for each $k \in \mathbb{N}$ let;

\begin{align} \quad E_{n, k} &= \bigcup_{k=1}^{\infty} \left \{ x \in E_n : f(x) > \frac{1}{k} \right \} \end{align}

- Note that $\{ x \in E_n : f(x) > 0 \} = \bigcup_{k=1}^{\infty} E_{n,k}$. Now we have that:

\begin{align} \quad \frac{1}{k} m(E_{n,k}) = \int_{E_{n,k}} \frac{1}{k} \leq \int_{E_{n,k}} f = 0 \end{align}

- Therefore $m(E_{n,k}) = 0$ for all $n \in \mathbb{N}$ and for all $k \in \mathbb{N}$. Hence $m(\{ x \in E_n : f(x) > 0 \}) = 0$ for all $n \in \mathbb{N}$ and so $f(x) = 0$ almost everywhere on $[a, b]$. $\blacksquare$

Lemma 2: Let $f$ is a Lebesgue integrable function on $[a, b]$ and let $\displaystyle{F(x) = \int_a^x f(t) \: dt}$. If for all $x \in [a, b]$ we have that $F(x) = 0$ then $f(x) = 0$ almost everywhere on $[a, b]$. |

**Proof:**Let $c, d \in [a, b]$ be such that $a \leq c \leq d \leq b$. Then:

\begin{align} \quad F(c) = \int_a^c f(t) \: dt = 0 \quad \mathrm{and} \quad F(d) = \int_a^d f(t) \: dt = 0 \end{align}

- So for every open subinterval $(c, d)$ of $[a, b]$ we have that:

\begin{align} \quad F(d) - F(c) = \int_c^d f(t) \: dt = 0 \quad (*) \end{align}

- Let $O$ be any open subset of $(a, b)$. Then we can express $O$ as a countable disjoint union of open intervals, say $\displaystyle{O = \bigcup_{k=1}^{\infty} (c_k, d_k)}$. So for every open subset $O$ of $[a, b]$ we have that:

\begin{align} \quad \int_O f(t) \: dt = \int_{\bigcup_{k=1}^{\infty} (c_k, d_k)} f(t) \: dt = \sum_{k=1}^{\infty} \int_{c_k}^{d_k} f(t) \: dt \overset{(*)} = \sum_{k=1}^{\infty} 0 = 0 \quad (**) \end{align}

- Let $S = \{ x \in (a, b) : f(x) > 0 \}$ and suppose that $m(S) > 0$. Then for $\displaystyle{\epsilon^* = \frac{1}{2}m(S) > 0}$ there exists a closed Lebesgue measurable subset $F \subseteq S$ such that $\displaystyle{m (S \setminus F) < \epsilon^* = \frac{1}{2}m(S)}$. By the excision property of the Lebesgue measure we have that:

\begin{align} \quad m(S) - m(F) < \frac{1}{2} m(S) \quad \Leftrightarrow \quad \frac{1}{2} m(S) < m(F) \end{align}

- Therefore:

\begin{align} \quad 0 = \int_a^b f(t) \: dt = \int_F f(t) \: dt + \underbrace{\int_{(a, b) \setminus F} f(t) \: dt}_{=0 \: \mathrm{by} \: (**) \: \mathrm{since \:} (a, b) \setminus F \: \mathrm{is \: open}} = \int_F f(t) \: dt \end{align}

- Therefore $\displaystyle{\int_F f(t) \: dt = 0}$. We have that $f$ is nonnegative on $F$, so by the previous lemma we have that $f(x) = 0$ almost everywhere on $F$. But this is a contradiction since $F \subseteq S$ and $f$ is positive on all of $S$. So the assumption that $m(S) > 0$ was false. So $m(S) = 0$. That is, $f(x) = 0$ almost everywhere on $[a, b]$. $\blacksquare$