TheoremBase

Bounded Measurable Functions on a Finite Measure Space Are Integrable

On a finite measure space a bounded measurable real function is integrable, and adding its bound gives a nonnegative integrable function whose integral is the integral of the function plus the bound times the total mass.

Statement

In the setting of Measurable Functions and the Lebesgue Integral: Notation, let (X,F,μ)(X,\mathcal{F},\mu) be a measure space whose measure μ\mu is finite, that is, μ(X)<∞\mu(X)<\infty. Let f:X→Rf:X\to\mathbb{R} be measurable, and let B∈RB\in\mathbb{R} with ∣f(x)∣≤B|f(x)|\le B for every x∈Xx\in X. Write f+B:X→Rf+B:X\to\mathbb{R} for the pointwise sum of ff and the constant function with value BB, so that

(f+B)(x)=f(x)+B(x∈X).(f+B)(x)=f(x)+B\qquad(x\in X).

Integrability. μ(X)\mu(X) is a real number, and f∈L1(μ)f\in\mathcal{L}^{1}(\mu).

The shifted function. The function f+Bf+B lies in M+\mathcal{M}^{+} and in L1(μ)\mathcal{L}^{1}(\mu), its integral as an element of M+\mathcal{M}^{+} equals its integral as an element of L1(μ)\mathcal{L}^{1}(\mu), and

∫X(f+B) dμ=∫Xf dμ+B μ(X).\int_{X}(f+B)\,d\mu=\int_{X}f\,d\mu+B\,\mu(X).

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