Show that the Riemann integration itself a continuous function

Question

Consider the function from cts$([a,b],\mathbb{R}) \rightarrow \mathbb{R}$ with $f \mapsto \int_{[a,b]}f$ where $\int$ denotes the Riemann integral. I want to prove that it is continuous.

Intuitively, I think that is true because considering the compact-open topology on the function space, if we varies the function by some small distance (under $d_\infty$), the integral will also varies by a small amount.

I think I can make use of complete metric space and the fact that if whenever a sequence of functions $f_n \rightarrow f$, $\int f_n \rightarrow\int f$ such that the Riemann integration will be continuous.

However, I am struggling on writing a formal statement about this. Would anyone give me some help? Thanks!

Answer 1

Using the $\|.\|_{\infty}$ norm on $\mathcal{C}^0([a,b])$, the linear function $I : f \mapsto \int_a^b f$ satisfies \begin{align} |I(f)| \leqslant \| f\|_{\infty}(b-a) \end{align} Hence it is continuous with respect to the topology induced by this norm.

Answer 2

Suppose $f\in cts([a,b],\mathbb{R})$, since $f$ is continuous, it is Riemann integrable and hence we can write $$\int_{[a,b]}f=\sum_{i=0}^{n-1}(t_{i+1}-t_i)x_i$$ where the $t$‘s represent the partition of $[a,b]$ and for some $x_i\in\mathbb{R}$. Then we have $$|\int_{[a,b]}f| = |\sum_{i=0}^{n-1}(t_{i+1}-t_i)x_i| \leq \sum_{i=0}^{n-1}(t_{i+1}-t_i)|x_i| \leq \sum_{i=0}^{n-1}(t_{i+1}-t_i)\;sup\{|f(x)|:x\in[a,b]\} = (b-a)sup\{|f(x)|:x\in[a,b]\}$$ which is the bound we wanted to use to prove the statement.

Now, let $f,g\in cts([a,b],\mathbb{R})$. Since the Riemann integral is linear, and since $f-g$ is continuous for any $f,g$ continuous, we have $$|\int_{[a,b]}f-\int_{[a,b]}g| = |\int_{[a,b]}f-g| \leq (b-a)\;sup\{|f(x)-g(x)|:x\in[a,b]\} = (b-a)\;d_\infty(f,g)$$ hence, let $\epsilon > 0$, if we let $\delta = \frac{\epsilon}{(b-a)}$ we have $$d_\infty(f,g)<\delta\Rightarrow |\int_{[a,b]}f-\int_{[a,b]}g|<\epsilon$$ and therefore the Riemann integral is continuous.