Prove that the map $f:[0,1]\to l_{\infty}([0,1];\Bbb{R})$ defined by $t\mapsto 1_{[t,1]}$ is Riemann integrable.

Question

A map $f:[0,1]\to F$ (where $F$ is a banach space) is said to be Riemann integrable if $\exists I\in F$ such that for every $\epsilon>0$ there is $\delta>0$ satisfying $$\left\lVert I-\sum\limits_{i=1}^n f(\xi_i)(t_i-t_{i-1})\right\rVert_F<\epsilon$$ for all partions $P=\{0=t_0<t_1<\cdots<t_n=1\}$ of $[0,1]$ with $\lVert P\rVert:=\underset{i}{\text{max}}(t_i-t_{i-1})<\delta$ and for every choice of $\xi_i\in(t_{i-1},t_i)$

Here $l_{\infty}([0,1];\Bbb{R}):=\left\{f:[0,1]\to\Bbb{R}|\ \lVert f\rVert_\infty=\underset{t}{\text{sup}}| f(t)|<\infty\right\}$ is banach space with the sup norm $\lVert\cdot\rVert_\infty$.

The map $f:[0,1]\to l_{\infty}([0,1];\Bbb{R})$ is defined as $\displaystyle{f(t)=1_{[t,1]}}\ \forall t\in[0,1]$. Recall that- $$ 1_{[t,1]}(x)=\begin{cases} 1&\text{if }x\in [t,1]\\ 0&\text{if }x\in [0,t) \end{cases}$$ I have to find $I\in l_{\infty}([0,1];\Bbb{R})$ such that the expression above in the definition of riemann integrebility is satisfied.

I don't have any idea how to think about this $I$. Initially, I have tried with $I=1_{[0,1]}$ (which is basically the constant function $1$), but it will not work because for any partion $P=\{0=t_0<t_1<\cdots<t_n=1\}$ we have $\left| 1_{[0,1]}(0)-\sum\limits_{i=1}^n f(\xi_i)(0)(t_i-t_{i-1})\right|=\left| 1-1_{[\xi_1,1]}(0)t_1\right|=1$ (as $\xi_1>0$).

This implies $\left\lVert 1_{[0,1]}-\sum\limits_{i=1}^n f(\xi_i)(t_i-t_{i-1})\right\rVert_\infty\ge 1$. So, $I=1_{[0,1]}$ fails.

Then I tried with constant function $0$, but it fails in the similar fashion (by evaluating at $1$).

Can anyone help me in this regard? Thanks for help in advance.

Answer 1

For partition $P$, the Riemann sum is the function $S:x \mapsto S(x)$ given by

$$S(x) = \sum_{j=1}^n \mathbf{1}_{[\xi_j,1]}(x)(t_j - t_{j-1}) = \sum_{j=1}^{n-1} t_j\mathbf{1}_{[\xi_j,\xi_{j+1})}(x) + \mathbf{1}_{[\xi_n,1]}(x)$$

As the partition norm tends to $0$, it appears that this sum -- which itself is a step function with respect to $x$ -- will converge to the integral $I:x \mapsto I(x) = x$.

To verify, note that if $x \in [0,1]$ then one of three cases holds:

(1) We have $x < \xi_1$ and $|S(x) - x| = x \leqslant t_1-0 \leqslant \|P\|$.

(2) We have $x \geqslant \xi_n$ and $|S(x) - x| = 1-x\leqslant 1-t_{n-1} \leqslant \|P\|$.

(3) We have $\xi_k \leqslant x < \xi_{k+1}$ for some $k \in \{1,2,\ldots,n-1\}$ and $$|S(x) - x|= |x - t_k| \leqslant \max(|t_{k+1}-t_k|, |t_k-t_{k-1}|)\leqslant \|P\|$$

Thus, $S(x) \to I(x) = x$ as $\|P\| \to 0$.