integral of sum of a sequence is sum of the integrals

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Theorem

Let ${f_{n}}_{n}$ be a sequence in $L^{+} (E)$ . Then

\int_{E} \sum_{n} f_{n} = \sum_{n} \int_{E} f_{n}

Proof

via induction. Since the lebesgue integral of sum is sum of the integral, we know for each $N \in N$ that

\int_{E} \sum_{n = 1}^{N} f_{n} = \sum_{n = 1}^{N} \int_{E} f_{n}

And since

\sum_{n = 1}^{1} f_{n} \leq \sum_{n = 1}^{2} f_{n} \leq \dots

and $\sum_{n = 1}^{N} f_{n} \to \sum_{n = 1}^{\infty} f_{n}$ as $N \to \infty$ , by the Monotone Convergence Theorem we have

\begin{aligned} \int_{E} \sum_{n = 1}^{\infty} & = lim_{N \to \infty} \int_{E} \sum_{n = 1}^{N} f_{n} \\ = lim_{N \to \infty} \sum_{n = 1}^{N} \int_{E} f_{n} \\ = \sum_{n = 1}^{\infty} \int_{E} f_{n} \end{aligned}

\begin{matrix} ◼ \end{matrix}

Note

this does not hold for Riemann integration. Enumerate the rationals and let $f_{n}$ be the function that is 1 for the first $n$ rational numbers and 0 everywhere else

integral of sum of a sequence is sum of the integrals

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