Integrals of Nonnegative Measurable Functions that Equal Zero

# Integrals of Nonnegative Measurable Functions that Equal Zero

Theorem 1: Let $(X, \mathcal A, \mu)$ be a complete measure space. If $f$ is a nonnegative measurable function defined on a measurable set $E$ and $\displaystyle{\int_E f(x) \: d \mu = 0}$ then $f(x) = 0$ $mu$-almost everywhere on $E$. |

**Proof:**Let $f$ be a nonnegative measurable function defined on a measurable set $E$. By The Simple Function Approximation Lemma and Theorem for General Measurable Spaces there exists a pointwise increasing sequence $(\varphi_n(x))_{n=1}^{\infty}$ of nonnegative simple functions that converges pointwise to $f(x)$. So for all $n \in \mathbb{N}$ we have that:

\begin{align} \quad 0 \leq \int_E \varphi_n(x) \: d \mu \leq \int_E f(x) \: d \mu = 0 \end{align}

- So for all $n \in \mathbb{N}$ we have that:

\begin{align} \quad \int_E \varphi_n(x) \: d \mu = 0 \end{align}

- Let $a_1, a_2, ..., a_m > 0$ be the positive real numbers in the range of $\varphi_n(x)$ such that $\varphi_n(x) = a_k$ if and only if $x \in E_k \subseteq E$ and:

\begin{align} \quad \int_E \varphi_n(x) \: d \mu = \sum_{k=1}^{m} a_k \mu (E_k) = 0 \end{align}

- So $a_k \mu (E_k) = 0$ for all $k \in \{ 1, 2, ..., m \}$. So $\mu (E_k) = 0$ for all $k \in \{ 1, 2, ..., m \}$. So $\mu (E_1 \cup E_2 \cup ... \cup E_m) = 0$ and $\varphi_n(x) > 0$ if and only if $x \in E_1 \cup E_2 \cup ... \cup E_m$.

- So $\varphi_n(x) = 0$ $mu$-almost everywhere on $E$.

- Let $E_0 = \bigcup_{n=1}^{\infty} \{ x \in E : \varphi_n(x) > 0 \}$. Then from the above argument we have that $\mu (E_0) = 0$ and $f(x) > 0$ if and only if $x \in E_0$. So $f(x) = 0$ $\mu$-almost everywhere on $E$. $\blacksquare$