PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31

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Null functions form a linear subspace and are exactly the zero-seminorm class

Statement

Let $(X, A, μ)$ be a measure space.

For each $1 \leq p < \infty$ , the set $N_{p} (μ) : = L^{p} (μ) \cap N (μ)$ is a linear subspace of $L^{p} (μ)$ , and for $f \in L^{p} (μ)$ one has $f \in N (μ) ⟺ ∥ f ∥_{p} = 0.$
The set $N_{\infty} (μ) : = L^{\infty} (μ) \cap N (μ)$ is a linear subspace of $L^{\infty} (μ)$ , and for $f \in L^{\infty} (μ)$ one has $f \in N (μ) ⟺ ∥ f ∥_{\infty} = 0.$

Facts & Assumptions

Given: A measure space $(X, A, μ)$ .

[L1]

The null functions are those that vanish almost everywhere (The null subspace of measurable functions that vanish almost everywhere).

[L2]

$L^{p} (μ)$ and $L^{\infty} (μ)$ are vector spaces in the relevant ranges ( $L^{p}$ and $L^{\infty}$ are vector spaces for $p \geq 1$ ).

[L3]

A countable union of measurable null sets is null (Finite and countable subadditivity of measures).

[L4]

A linear subspace means the three closure conditions of Linear subspace of a vector space.

[L5]

For $1 \leq p < \infty$ , a nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere).

[L6]

If $∥ f ∥_{\infty} < \infty$ , then $∣ f ∣ \leq ∥ f ∥_{\infty}$ almost everywhere (The essential supremum is attained as the least essential bound).

Proof

technique · Use countable-union stability of null sets for addition and scalar multiplication. For $1 \le p < infinity$, the $p$-seminorm vanishes exactly when the integral of $|f|^p$ is zero; for $p = infinity$, vanishing means the essential supremum is zero

1.1L1L2L3L4

Fix $1 \leq p < \infty$ . If $f, g \in N_{p} (μ)$ , choose measurable null sets $E_{f}, E_{g}$ outside which $f = 0$ and $g = 0$ . Their union is null, and on its complement one has $f + g = 0$ and $a f = 0$ for every $a \in R$ . Because $L^{p} (μ)$ is a vector space, [L4] makes $N_{p} (μ)$ a linear subspace.

1.2

If $f \in N_{p} (μ)$ , then $∣ f ∣^{p} = 0$ almost everywhere, so [L1, L5] $\int ∣ f ∣^{p} d μ = 0,$ hence $∥ f ∥_{p} = 0$ . Conversely, if $∥ f ∥_{p} = 0$ , then the same theorem [L5] forces $∣ f ∣^{p} = 0$ almost everywhere and therefore $f = 0$ almost everywhere.

1.3L1L6

If $f \in N_{\infty} (μ)$ , then $0$ is an essential bound for $∣ f ∣$ , so $∥ f ∥_{\infty} = 0$ . Conversely, if $∥ f ∥_{\infty} = 0$ , then [L6] gives $∣ f ∣ \leq 0$ almost everywhere, hence $f = 0$ almost everywhere.

2.1L1L2L3L4

If $f, g \in N_{\infty} (μ)$ , the same null-set union argument as in step 1.1 shows that $f + g$ and $a f$ vanish almost everywhere, and [L2] places them in $L^{\infty} (μ)$ . Therefore [L4] makes $N_{\infty} (μ)$ a linear subspace.

3.1step 1.1step 1.2step 2.1step 1.3∎

Steps 1.1 and 2.1 prove the two subspace claims, and steps 1.2 and 1.3 identify the zero-seminorm class in every range.

Null functions form a linear subspace and are exactly the zero-seminorm class

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources