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So I have $f_n:[a,b] \rightarrow \mathbb{R}$ is continuous s.t. $f_n \rightarrow f$ uniformly on $[a,b]$. I need to show that $$ \lim_{n \rightarrow \infty} \int^b_a f_n = \int^b_a f$$

My attempt at doing that is:

If we have that $f$ is bounded, then by uniform convergence (where $\epsilon =1$) there exists a $n_0$ s.t. $$ |f_{n_0}(x) - f(x)| < 1$$ $\Rightarrow |f(x)| \leq |f_{n_0}|+1 < M$ for all $x \in [a,b]$ and some M.

In addition now let $\epsilon > 0$. By uniform convergence it must be true that there exists N such that $|f_n(x)-f(x)|<\epsilon$ for all $x \in [a,b]$ and all $n \geq N$.

We can then use Riemann's condition, my logic being that because $f_n$ is continuous then it also follows that it is integrable? If this is the case then there exists a dissection $D$ of $[a,b]$ s.t:

$U(f_n(D))-L(f_n(D)) < \epsilon(b-a)$

Which are the upper and lower sums of $f_n$ respectively. Now we can prove that $U(f(D))-L(f(D))$ is small too.

Since we know that $f \leq f_N + \epsilon$ it implies that $ \sup \ f[x_{i-1}, x_i] \leq \sup \ f_N[x_{i-1},x_i] + \epsilon$ $$\Rightarrow U(f(D))\leq L(f_n(D)) + \epsilon (b-a)$$ $$\Rightarrow U(f(D))\geq L(f_n(D)) - \epsilon (b-a)$$

Subtracting the two above inequalities results in $$U(f(D))-L(f(D)) < U(f_n(D))-L(f_n(D)) + 2\epsilon (b-a) \leq 3 \epsilon (b-a) $$

$\epsilon$ is arbitrary and therefore by Riemann's condition, $f$ is integrable.

Thus we can close our argument with: $$ \left| \int^b_a f - \int^b_a f_N\right| = \left|\int^b_a (f-f_N) \right| \leq \int^b_a |f-f_N| \leq \epsilon(b-a)$$

$\epsilon$ is arbitrary again so our result follows?

That is the proof I believe it to be but some of my classmates aren't convinced. Any help would be appreciated.

sai-kartik
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Pierre Bourne
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1 Answers1

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It looks correct, but there is no need to work with upper and lower sums here.

Note that, since each $f_n$ is continuous and $(f_n)_{n\in\Bbb N}$ converges uniformly to $f$, then $f$ is continuous too, and therefore Riemann-integrable.

Now, given $\varepsilon>0$, take $N\in\Bbb N$ such that$$n\geqslant N\implies\sup_{x\in[a,b]}\bigl|f(x)-f_n(x)\bigr|<\frac\varepsilon{b-a}.$$Then, if $n\geqslant N$,\begin{align}\left|\int_a^bf(x)\,\mathrm dx-\int_a^bf_n(x)\,\mathrm dx\right|&=\left|\int_a^bf(x)-f_n(x)\,\mathrm dx\right|\\&\leqslant\int_a^b\bigl|f(x)-f_n(x)\bigr|\,\mathrm dx\\&<(b-a)\frac\varepsilon{b-a}\\&=\varepsilon.\end{align}

José Carlos Santos
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