Definite Integral of Limit of Uniformly Convergent Sequence of Integrable Functions
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Theorem
Let $a, b \in \R$ with $a < b$.
Let $\sequence {f_n}$ be a sequence of Riemann integrable real functions $\closedint a b \to \R$ converging uniformly to $f : \closedint a b \to \R$.
Then $f$ is integrable, and:
- $\ds \int_a^b \map f x \rd x = \lim_{n \mathop \to \infty} \int_a^b \map {f_n} x \rd x$
Proof
By Limit of Uniformly Convergent Sequence of Integrable Functions is Integrable, $f$ is integrable.
We have:
\(\ds \size {\int_a^b \map f x \rd x - \int_a^b \map {f_n} x \rd x}\) | \(=\) | \(\ds \size {\int_a^b \paren {\map f x - \map {f_n} x} \rd x}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \int_a^b \size {\map f x - \map {f_n} x} \rd x\) | Triangle Inequality for Integrals | |||||||||||
\(\ds \) | \(\le\) | \(\ds \paren {b - a} \sup_{x \mathop \in \closedint a b} \size {\map f x - \map {f_n} x}\) | Darboux's Theorem |
Let $\epsilon \in \R_{> 0}$.
Since $f_n \to f$ uniformly, we can find $N \in \N$ such that for $n > N$ we have:
- $\ds \sup_{x \mathop \in \closedint a b} \size {\map f x - \map {f_n} x} < \frac \epsilon {b - a}$
by the definition of uniform convergence.
So, for $n > N$ we have:
- $\ds \paren {b - a} \sup_{x \mathop \in \closedint a b} \size {\map f x - \map {f_n} x} < \epsilon$
and hence:
- $\ds \size {\int_a^b \map f x \rd x - \int_a^b \map {f_n} x \rd x} < \epsilon$
Since $\epsilon$ was arbitrary it follows that:
- $\ds \lim_{n \mathop \to \infty} \int_a^b \map {f_n} x \rd x = \int_a^b \map f x \rd x$
as required.
$\blacksquare$