lebesgue integral of sum is sum of the integral

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Theorem

If $f, g \in L^{+} (E)$ then

\int_{E} (f + g) = \int_{E} f + \int_{E} g

Proof

Let ${φ_{n}}_{n}$ and ${ψ_{n}}_{n}$ be two sequences of simple functions such that
$0 \leq φ_{1} \leq φ_{2} \leq \dots \leq f$ with $φ_{n} \to f$ and
$o \leq ψ_{1} \leq ψ_{2} \leq \dots \leq g$ with $ψ_{n} \to g$ pointwise then

0 \leq φ_{1} + ψ_{1} \leq φ_{2} + ψ_{2} \leq \dots \leq f + g

Where $φ_{n} + ψ_{n} \to f + g$ pointwise and each $φ_{i} + ψ_{i}$ are simple (since they are the sum of simple functions). So by the Monotone Convergence Theorem and linearity for simple functions we have

\begin{aligned} \int_{E} (f + g) & = lim_{n \to \infty} \int_{E} (φ_{n} + ψ_{n}) \\ (linearity) & = lim_{n \to \infty} \int_{E} φ_{n} + \int_{E} ψ_{n} \\ (MTC) & = \int_{E} f + \int_{E} g \end{aligned}

References

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lebesgue integral of sum is sum of the integral

References

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