Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-01
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Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in Rm

Statement

Let Q=∏j<m[aj,bj] be nondegenerate. For integrable f,g:Q→R and scalars α,β, the function αf+βg is integrable and its integral is α∫Qf+β∫Qg. If f≤g, then ∫Qf≤∫Qg. Also ∣f∣ is integrable and ∣∫Qf∣≤∫Q∣f∣. If ar<c<br, cutting Q at the coordinate hyperplane xr=c gives two nondegenerate subrectangles; integrability on Q is equivalent to integrability on both restrictions, and their integral values add to the integral over Q.

Facts & Assumptions

Given: The stated integrable functions on the nondegenerate rectangle, and, for coordinate-slice additivity, a strictly interior cut ar<c<br.

[L3]

∣∣u∣−∣v∣∣≤∣u−v∣, the reverse triangle inequality on the real line (The reverse triangle inequality, Absolute value in an ordered field, Basic properties of the absolute value).

Proof

technique · direct
1.1L1L2

Refine grids good for f and g. Cellwise supremum and infimum estimates make the gap of αf+βg at most ∣α∣ times the gap of f plus ∣β∣ times that of g; tagged-sum linearity identifies the value.

1.2L1L3given

Termwise f≤g gives monotonicity of every tagged sum and hence of integrals. By [L3], the oscillation of ∣f∣ on a cell is no larger than that of f, so ∣f∣ is integrable; −∣f∣≤f≤∣f∣ then gives the absolute-value estimate.

1.3L1L2

Insert the cut coordinate into the grid. [L2] splits every Darboux or tagged sum into the two subrectangle sums. Good grids splice conversely, proving integrability on Q exactly when both restrictions are integrable, and proving additivity.

2.1step 1.1step 1.2step 1.3∎

These arguments establish all clauses with positively oriented rectangles.

Depends on

Dependency tree · 2 levels

Sources