# Proof critique

• Feb 3rd 2011, 03:47 PM
mathman88
Proof critique
Hello, I am trying to prove that given a proper integral, $\displaystyle \int_a^b \sum_{n=0}^\infty f_n(x) dx = \sum_{n=0}^\infty \int_a^b f_n(x) dx$, assuming both converge.

Here's my attempted proof:

$\displaystyle \sum_{n=0}^\infty f_n(x) = \sum_{n=0}^N f_n(x) + R_N(x)$, where $\forall x, \; R_N(x) \to 0$ as $N \to \infty$.

Thus $\displaystyle \int_a^b \sum_{n=0}^\infty f_n(x) dx = \int_a^b \sum_{n=0}^N f_n(x) dx + \int_a^b R_N(x) dx = \sum_{n=0}^N \int_a^b f_n(x) dx + \int_a^b R_N(x) dx$.

But $\displaystyle 0 \leq \left| \int_a^b R_N(x) dx \right| \leq \int_a^b \left|R_N(x)\right| dx \leq (a-b)\cdot\max_{a\leq x\leq b}\bigg\{|R_n(x)|\bigg\} \to 0$ as $N \to \infty$.

Hence $\displaystyle \int_a^b \sum_{n=0}^\infty f_n(x) dx = \sum_{n=0}^\infty \int_a^b f_n(x) dx$.

Does this look valid?
• Feb 3rd 2011, 09:22 PM
Jose27
Quote:

Originally Posted by mathman88
$(a-b)\cdot\max_{a\leq x\leq b}\bigg\{|R_n(x)|\bigg\} \to 0$ as $N \to \infty$.

There's a mistake in this step, since we don't know that we can bound $R_N(x)$ uniformly on the interval.

That said what you're trying to prove is false as stated (indeed, if it were true, theorems like dominated and monotone convergence would be unecessary), there is a sequence of functions such that all terms have the same integral and the sequence converges pointwise to zero.