Zero Integral and Almost-Everywhere Vanishing

What Does a Zero Integral Tell Us?

A nonnegative measurable function can be positive at some points and still have integral zero. This happens when its positive values are confined to a set of measure zero. The precise statement is an equivalence: zero integral means that the function vanishes almost everywhere, and vanishing almost everywhere forces the integral to be zero. The measurable sets on which the function exceeds a positive threshold make both directions transparent.

Throughout, let \((X,\mathcal{M},\mu)\) be an arbitrary measure space. No assumption that \(\mu\) is finite, sigma-finite, or complete is needed. The function \(f:X\to[0,\infty]\) is extended-real-valued and \(\mathcal{M}\)-measurable; its integral is the nonnegative Lebesgue integral, which is allowed to be \(+\infty\). A property holds \(\mu\)-almost everywhere (\(\mu\)-a.e.) if it fails only on a measurable set of \(\mu\)-measure zero. In particular, \(f=0\) \(\mu\)-a.e. means \(\mu(\{x:f(x)\ne0\})=0\). A measurable function need not be integrable: here, “integrable” for a nonnegative function means that its integral is finite.

Recall that if \(\psi=\sum_{j=1}^k a_j\chi_{A_j}\) is a nonnegative simple function with disjoint measurable sets \(A_j\) and finite coefficients \(a_j\ge0\), then \(\int\psi\,d\mu=\sum_{j=1}^k a_j\mu(A_j)\). The nonnegative integral of a measurable \(f\) is the supremum of the integrals of all nonnegative simple functions \(\psi\le f\). We use this definition and monotonicity of the nonnegative integral.

The Zero-Integral Criterion

Proof. First suppose \(f=0\) \(\mu\)-a.e. The set \(N=\{x:f(x)>0\}\) is measurable and has measure zero. Let \(\psi\) be any nonnegative simple function with \(\psi\le f\). If \(\psi(x)>0\), then \(f(x)>0\), so \(\psi\) vanishes outside \(N\). Write \(\psi=\sum_{j=1}^k a_j\chi_{A_j}\) using disjoint measurable sets and positive coefficients for its nonzero values. Every such \(A_j\) is contained in \(N\), so \(\mu(A_j)=0\) and \(\int\psi\,d\mu=\sum_{j=1}^k a_j\mu(A_j)=0\). The supremum over all such \(\psi\) is therefore zero, proving \(\int f\,d\mu=0\).

Conversely, suppose \(\int f\,d\mu=0\). For each integer \(n\ge1\), define the measurable strict level set \(E_n=\{x:f(x)>1/n\}\). The simple function \((1/n)\chi_{E_n}\) is at most \(f\), so monotonicity and the definition of the integral give \(\frac{1}{n}\mu(E_n)=\int (1/n)\chi_{E_n}\,d\mu\le\int f\,d\mu=0\). Since \(\mu(E_n)\ge0\), this forces \(\mu(E_n)=0\). Moreover, \(\{f>0\}=\bigcup_{n=1}^{\infty}E_n\): if \(f(x)>0\), including if \(f(x)=+\infty\), there is an integer \(n\) with \(1/n<f(x)\); the reverse inclusion follows from \(1/n>0\). Countable subadditivity now gives \(\mu(\{f>0\})=0\), which means \(f=0\) \(\mu\)-a.e. This proves both directions. \(\square\)

The proof uses level sets at thresholds \(1/n\) because every strictly positive value, however small, exceeds at least one such threshold. The argument also handles infinite values: if \(f(x)=+\infty\), then \(x\in E_n\) for every \(n\). It does not require \(\mu(X)<\infty\), and it does not require \(f\) to be bounded.

Proof. Suppose first that \(\int f\,d\mu=0\). The theorem gives \(\mu(\{f>0\})=0\), and \(\{f>t\}\subseteq\{f>0\}\) for every \(t>0\). Thus every positive level set has measure zero. Conversely, suppose \(\mu(\{f>t\})=0\) for every \(t>0\). In particular, each set \(\{f>1/n\}\) has measure zero. Their union is \(\{f>0\}\), so \(f=0\) \(\mu\)-a.e.; the theorem gives \(\int f\,d\mu=0\). \(\square\)

Worked Examples with Level Sets

Common Mistakes and Why the Criterion Matters

Confusing positive with positive almost everywhere. A zero integral does not imply that \(f(x)=0\) for every \(x\). In the second example \(f(b)=7\), but \(\{b\}\) is null, so \(f=0\) \(m\)-a.e. Conversely, if the measure of a point is positive, a positive value there can contribute a positive amount to the integral, as the counting-measure example demonstrates. The measure space matters: the same underlying set need not treat its subsets the same way under different measures.

Misreading a strict level-set inequality. A common incorrect calculation for \(f=\chi_{\mathbb{Q}\cap[0,1]}\) is to claim that \(\{f>1\}=\mathbb{Q}\cap[0,1]\). This is false: on the rationals in the interval \(f=1\), and \(1>1\) is false, so \(\{f>1\}=\varnothing\). The rational set is the level set \(\{f>1/n\}\) only for \(n\ge2\). Likewise, for \(f=7\chi_{\{b\}}\), the set \(\{f>7\}\) is empty, not \(\{b\}\); the singleton occurs when \(0<t<7\). Keeping the strict inequality explicit prevents endpoint errors.

The theorem is useful because it turns an integral statement into a measure statement about the support of positive values. One does not need to compute an integral by approximation in order to establish that a function is zero almost everywhere: it is enough to show that each positive level set is null. In the other direction, once the exceptional set is known to be null, the integral is forced to be zero. The theorem concerns nonnegative measurable functions, including those whose integral might initially be infinite; it does not assert a corresponding zero-integral criterion for arbitrary signed functions without additional definitions and hypotheses.