# Bounded convergence theorem for Riemann integrals

I will teach a analysis class to some olympiad students this month. The subject is the fundamental theorem of algebra. My approach will be to prove the following "modified" version of the bounded convergence theorem:

Let $\{f_n\}$ be a sequence of Riemann Integrable functions such that $f_n:[a,b]\rightarrow\mathbb{R}$ and $|f_n(x)|<M$ for all $n\geq1$ with $M>0$. Suppose that $f_n\to f$ pointwise where $f:[a,b]\rightarrow\mathbb{R}$ is Riemann Integrable. Then $$\lim\limits_{n\to\infty}\int_a^bf_n(x)\,\mathrm{d}x = \int_a^bf(x)\,\mathrm{d}x.$$

Then I will show that this implies in

Let $f(x,t)$ be integrable in $x$ for each $t$ and have a bounded partial derivative $f_t(x,t)$ which is integrable in $x$ for each $t$. Then $$\frac{\mathrm{d}}{\mathrm{d}t} \int_a^b f(x,t)\mathrm{d}x = \int_a^b f_t(x,t)\mathrm{d}x.$$

Last, but not least, I will show that this implies in the fundamental theorem of algebra.

The class knows a lot about elementary calculus but doesn't know anything about the Lebesgue integral. I want to know what is the best/easiest proof of this modified version of the bounded convergence theorem.

Thanks.

EDIT: In rough terms, the idea is as follows:

Let $P$ be a polynomial, define $g:[0,\infty)\times[0,2\pi]\to\mathbb{C}, \:\: g(r,\theta)=\displaystyle\frac{1}{P(re^{i\theta})}$ and $F:[0,\infty)\to\mathbb{C}, \quad F(r)=\int_0^{2\pi}g(r,\theta)\mathrm{d}\theta$. Suppose $P(z)\neq0$ for all $z\in\mathbb{C}$.

Differentiating $F$ under the integral sign (I want the bounded convergence theorem to justify this), we have that

$$riF'(r)=\int_0^{2\pi}\frac{-rie^{i\theta}P'(re^{i\theta})}{[P(re^{i\theta})]^2}\mathrm{d}\theta=\int_0^{2\pi}\frac{\partial g}{\partial \theta}(r,\theta)\mathrm{d}\theta=g(r,2\pi)-g(r,0)=0.$$

This implies $F(r)=\frac{2\pi}{P(0)}\neq0, \quad \forall r\geq0$.

On the other side, $|P(z)|\to\infty$ when $|z|\to\infty$ implies in $g(r,\theta)\to0$ when $r\to\infty$. Then $F(r)\to 0$ when $r\to\infty$, which is absurd.

I am having trouble justifying the use of the Leibniz Rule.

• The bounded convergence for Riemann integrals is an exercise in Rudin Real and Complex Analysis, with a comment that this is a good illustration of why the Lebesgue integral is a good thing, and a reference to a paper, I think in the Monthly, where it's proved. (Comment and reference may be in the comments at the end of the chapter - my copy of the book is missing...) – David C. Ullrich Jun 25 '16 at 15:54
• Surely you can get the application to FTA from something easier about the Riemann integral? For example it's easy to see that if $|f_n|\le M$ and $f_n\to f$ uniformly on $[a+\delta,b-\delta]$ for every $\delta>0$ then QED. – David C. Ullrich Jun 25 '16 at 15:56
• How is all this related to fundamental theorem of algebra? – Paramanand Singh Jun 25 '16 at 19:13
• I edited the post. – Gabriel Jun 25 '16 at 21:16
• @GabrielRibeiro Yes, but then you need the bounded convergence theorem, which is highly non-trivial! The Leibniz things for smooth functions in calculus books is enough here - the uniform convergence is not that hard, and then you're done. – David C. Ullrich Jun 28 '16 at 0:26
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