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Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree

Statement

Let $f, g \in L^{1} (μ)$ . Then the following are equivalent:

$f = g$ almost everywhere;
for every measurable $A$ , $\int_{A} f d μ = \int_{A} g d μ .$

For integrable real or complex $h$ , the notation in condition 2 means $\int_{A} h d μ : = \int h χ_{A} d μ$ ; the product is integrable because $∣ h χ_{A} ∣ \leq ∣ h ∣$ .

Facts & Assumptions

Given: Integrable functions $f, g \in L^{1} (μ)$ .

[L1]

The integral over a null set vanishes for nonnegative integrands (A nonnegative integral over a null set vanishes).

[L2]

The Lebesgue integral is linear on $L^{1} (μ)$ (The Lebesgue integral is linear on $L^{1} (μ)$ ).

[L3]

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).

[L4]

Real and imaginary parts of an integrable complex function are integrable (Integrable real and complex functions, and their integrals).

Proof

technique · direct

1.1L1L2L4given

Assume $f = g$ almost everywhere, with exceptional null set $N$ . For every measurable $A$ , the positive and negative parts of the real and imaginary components of $(f - g) χ_{A}$ are supported on $N$ . By [L1] their nonnegative integrals vanish, and [L2] and [L4] give $\int_{A} (f - g) d μ = 0$ , hence $\int_{A} f d μ = \int_{A} g d μ$ .

1.2L2L3L4given

Assume instead that $\int_{A} f d μ = \int_{A} g d μ$ for every measurable $A$ . Apply this to the real part $u : = Re (f - g)$ on $A_{+} : = {u > 0}$ and to $- u$ on $A_{-} : = {u < 0}$ . In each case the corresponding nonnegative integral is $0$ , so [L3] gives $u = 0$ almost everywhere. The same argument for $v : = Im (f - g)$ shows $v = 0$ almost everywhere. Hence $f = g$ almost everywhere.

2.1step 1.1step 1.2∎

Step 1.1 proves $(1) \Rightarrow (2)$ and step 1.2 proves $(2) \Rightarrow (1)$ .

Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources