Definite Integral of Limit of Uniformly Convergent Sequence of Integrable Functions

From ProofWiki
Jump to navigation Jump to search

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:

$\displaystyle \int_a^b \map f x \rd x = \lim_{n \to \infty} \int_a^b \map {f_n} x \rd x$


Proof

By Limit of Uniformly Convergent Sequence of Integrable Functions is Integrable, $f$ is integrable.

We have:

\(\displaystyle \size {\int_a^b \map f x \rd x - \int_a^b \map {f_n} x \rd x}\) \(=\) \(\displaystyle \size {\int_a^b \paren {\map f x - \map {f_n} x} \rd x}\)
\(\displaystyle \) \(\le\) \(\displaystyle \int_a^b \size {\map f x - \map {f_n} x} \rd x\) Triangle Inequality for Integrals
\(\displaystyle \) \(\le\) \(\displaystyle \paren {b - a} \sup_{x \in \closedint a b} \size {\map f x - \map {f_n} x}\) Upper and Lower Bounds of Integral

Let $\varepsilon \in \R_{> 0}$.

Since $f_n \to f$ uniformly, we can find $N \in \N$ such that for $n > N$ we have:

$\displaystyle \sup_{x \in \closedint a b} \size {\map f x - \map {f_n} x} < \frac \varepsilon {b - a}$

by the definition of uniform convergence.

So, for $n > N$ we have:

$\displaystyle \paren {b - a} \sup_{x \in \closedint a b} \size {\map f x - \map {f_n} x} < \varepsilon$

and hence:

$\displaystyle \size {\int_a^b \map f x \rd x - \int_a^b \map {f_n} x \rd x} < \varepsilon$

Since $\varepsilon$ was arbitrary it follows that:

$\displaystyle \lim_{n \to \infty} \int_a^b \map {f_n} x \rd x = \int_a^b \map f x \rd x$

as required.

$\blacksquare$