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TheoremStatement: Literature-sourcedProof: AI-generatedverified 2026-09-26 (gpt-6-sol)
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Chebyshev-Markov inequality for the integral

Statement

Let f:X→[0,+∞] be measurable and let t>0. Then μ({f≥t})≤1t∫f dμ.

Facts & Assumptions

Given: A nonnegative measurable function f and a real number t>0.

[L1]

The nonnegative integral is monotone and homogeneous (Monotonicity and nonnegative homogeneity of the nonnegative integral).

[L2]

For measurable E and t>0, the nonnegative integral of the simple function tχE is tμ(E) (The nonnegative integral agrees with the simple integral on simple functions, The integral of a nonnegative simple function).

Proof

technique · direct
1.1L1L2given

Put E:={f≥t}, which is measurable. Since tχE≤f, [L1] and [L2] give t μ(E)=∫tχE dμ≤∫f dμ.

2.1step 1.1algebra∎

Dividing by the positive real t gives μ({f≥t})≤t−1∫f dμ.

Depends on

Used by

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources