Reconstructing the Riemann Integral Using the Completion of Normed Spaces

Question

Every normed space can be completed into a Banach space, and this completion satisfies a universal property:

Let $X$ be a normed space, $\tilde{X}$ its completion, and $c: X \to \tilde{X}$ the isometric embedding. For any Banach space $Y$ and any bounded linear map $T: X \to Y$, there exists a unique bounded linear map $\tilde{T}: \tilde{X} \to Y$ such that $\tilde{T}|_X = T$, $\|\tilde{T}\| = \|T\|$, and $T = \tilde{T} \circ c$.

I want to use this theorem to redefine the Riemann integral. Let $S[a, b]$ be the space of step functions on $[a, b]$, i.e., functions that are linear combinations of characteristic functions on intervals defined by a partition of $[a, b]$. Equip $S[a, b]$ with the $\sup$ norm. For example, $s \in S[a, b]$ can be written as: $$ s = \sum_{i=1}^n s_i \chi_i, $$ where $\chi_i$ is the characteristic function of the $i$-th interval $[x_{i-1}, x_i)$, with the last interval being closed if $i = n$. Define the integral $I: S[a, b] \to \mathbb{R}$ as a bounded linear map: $$ I\left(\sum_{i=1}^n s_i \chi_i\right) := \sum_{i=1}^n s_i (x_i - x_{i-1}). $$ Using the universal property, we can extend $I$ to the completion $\tilde{S}[a, b]$. I hope to achieve the following results, but I am unsure how to verify them:

1. $\tilde{S}[a, b]$ can be isometrically embedded into $L^{\infty}[a, b]$.
2. The space of Riemann integrable functions $R[a, b]$ is a subspace of $\tilde{S}[a, b]$, and the restriction of $\tilde{I}$ to $R[a, b]$ coincides with the Riemann integral.

Could someone help me with this? Thank you!

Answer 1

Since this answer is a bit long, here is a summary. The completion $\tilde{S}[a,b]$, realised as the closure of $S[a,b]$ in the Banach space of bounded functions on $[a,b]$, is a proper subspace of the space of regulated functions. The completion is explicitly characterised in the answer below and it is shown that this completion contains the continuous functions. Moreover, in relation to your proposed results, we have the following.

1. The completion isometrically embeds into $L^{\infty}[a,b]$.

2. The completion is a proper subspace of the Riemann integrable functions on $[a,b]$. Furthermore, the Riemann integral coincides with $\tilde{I}$ on $\tilde{S}[a,b]$.

We first establish the completion of $S[a,b]$ as a closed subspace of $B[a,b]$, where $B[a,b]$ denotes the Banach space of bounded (scalar-valued) functions on $[a,b]$. To achieve this, we use the following lemma.

Lemma. Let $f\colon [a,b] \to \mathbb{K}$. Then the following are equivalent.

$\text{(I)}$. The function $f$ has the following properties.

1. The function $f$ is right continuous at every point on $[a,b)$, that is \begin{equation} \lim_{x\downarrow c} f(x) = f(\alpha ) \end{equation} for all $c \in [a, b)$.

2. The limit \begin{equation} \lim_{x\uparrow c} f(x) \end{equation} exists for all $c \in (a, b)$.

3. The function $f$ is left continuous at $b$, that is \begin{equation} \lim_{x\uparrow b} f(x) = f(b) . \end{equation}

$\text{(II)}$. The function $f$ belongs to the uniform closure of $S[a,b]$ in $B[a,b]$.

Proof. $\text{(I)} \Rightarrow \text{(II)}$. Suppose $f$ is right continuous at every point on $[a,b)$, is left continuous at $b$ and that the limit \begin{equation} \lim_{x\uparrow c} f(x) \end{equation} exists for all $c \in (a,b)$. Let $\varepsilon > 0$. Since \begin{equation} \lim_{x\downarrow c} f(x) \end{equation} exists for all $c\in [a,b)$ and \begin{equation} \lim_{x\uparrow c} f(x) \end{equation} exist for all $c \in (a,b]$, we find for each $x\in [a,b]$ values $\alpha (x), \beta (x) \in \mathbb{R}$ with the following properties.

1. We have $\alpha (x) < x < \beta (x)$.

2. If $x\in [a,b)$ then for all $s,t\in (x, \beta (x)) \cap [a,b]$ we have $|f(s) - f(t)| \leq \varepsilon$.

3. If $x\in (a,b]$ then for all $s,t\in (\alpha (x), x) \cap [a,b]$ we have $|f(s) - f(t)| \leq \varepsilon$.

By compactness there are $x_{1}, \ldots , x_{n} \in [a,b]$ such that $[a,b] \subseteq \bigcup_{k=1}^{n} (\alpha (x_{k}), \beta (x_{k}))$. Let $(\xi_{0} , \ldots , \xi_{\ell})$ be the partition of $[a,b]$ such that \begin{equation} \{\xi_{0}, \ldots , \xi_{\ell} \} = [a,b] \cap \{a,b, \alpha (x_{1}), \ldots , \alpha (x_{n}), \beta (x_{1}), \ldots , \beta (x_{n})\} . \end{equation} Then for all $k\in \{1, \ldots , \ell\}$ and $s,t\in (\xi_{k-1}, \xi_{k})$ we have $|f(s) - f(t)| \leq \varepsilon$. By using that $f$ is right continuous at every point on $[a,b)$ and left continuous at $b$, we obtain the following.

1. For each $k\in \{1, \ldots , \ell - 1\}$ we have $|f(s) - f(t)| \leq \varepsilon$ for all $s,t\in [\xi_{k-1}, \xi_{k})$.

2. We have $|f(s) - f(t)| \leq \varepsilon$ for all $s,t\in [\xi_{\ell - 1}, \xi_{\ell} ]$.

Now define $g := \sum_{k=1}^{\ell - 1} f(\xi_{k-1}) 1_{[\xi_{k-1}, \xi_{k})} + f(\xi_{\ell - 1}) 1_{[\xi_{\ell - 1}, \xi_{\ell}]}$. Then $g\in S[a,b]$ and it follows from what we have already shown that $\|g - f\|_{\infty} \leq \varepsilon$. This shows $\text{(II)}$.

$\text{(II)} \Rightarrow \text{(I)}$. Suppose that $f$ belongs to the uniform closure of $S[a,b]$ in $B[a,b]$. It is a straightforward exercise to show the following. (The proof of Theorem VI.1.2 in Analysis II by Amann and Escher may help if needed.)

1. For every $c\in [a,b)$ the subspace \begin{equation} M_{1,c} := \{g\in B[a,b] : \lim_{x\downarrow c} g(x) \, \text{exists} \} \end{equation} is closed in $B[a,b]$.

2. For every $c\in [a,b)$ the subspace \begin{equation} M_{2,c} := \{g\in B[a,b] : \lim_{x\downarrow c} g(x) = g(c) \} \end{equation} is closed in $B[a,b]$.

3. For every $c\in (a,b]$ the subspace \begin{equation} M_{3,c} := \{g\in B[a,b] : \lim_{x\uparrow c} g(x) \, \text{exists} \} \end{equation} is closed in $B[a,b]$.

4. For every $c\in (a,b]$ the subspace \begin{equation} M_{4,c} := \{g\in B[a,b] : \lim_{x\uparrow c} g(x) = g(c) \} \end{equation} is closed in $B[a,b]$.

Note that each member of $S[a,b]$ belongs to the set \begin{equation} (\bigcap_{c\in [a,b)} M_{2,c}) \cap (\bigcap_{c\in (a,b)} M_{3,c}) \cap M_{4,b} , \tag{1} \end{equation} which is a closed subspace of $B[a,b]$ by what was stated above. Since $f$ belongs to the closure of $S[a,b]$, it follows that $f$ belongs to the set $(1)$. This shows $\text{(I)}$ and completes the proof.

As a consequence of the previous lemma, from now on we take $\tilde{S}[a,b]$ as the uniform closure of $S[a,b]$ in $B[a,b]$. Note that $\tilde{S}[a,b]$ contains all the continuous functions on $[a,b]$.

At this point it may be helpful to relate all of this to a well-known theory of integration. It will help to have some definitions. Let $f\colon [a,b] \to \mathbb{K}$ be a function. The function $f$ is called a staircase function if there is a partition $(\xi_{0}, \ldots , \xi_{n})$ of the interval $[a,b]$ (so that $\xi_{0} = a$, $\xi_{n} = b$ and $\xi_{k-1} < \xi_{k}$ for all $k\in \{1, \ldots , n\}$) such that $f\vert_{(\xi_{k-1}, \xi_{k})}$ is constant for all $k\in \{1, \ldots , n\}$. The function $f$ is called a jump continuous function if the limit \begin{equation} \lim_{x\downarrow c} f(x) \end{equation} exists for all $c\in [a,b)$ and the limit \begin{equation} \lim_{x\uparrow c} f(x) \end{equation} exists for all $c \in (a, b]$. Finally, the function $f$ is called a regulated function if $f$ is a uniform limit of a sequence of staircase functions.

The collections of staircase functions, jump continuous functions and regulated functions are all contained in $B[a,b]$. In particular, the space of regulated functions is the uniform closure of the space of staircase functions in $B[a,b]$ and the space of regulated functions equipped with the supremum norm is a Banach space. Moreover, a function is jump continuous if and only if it is regulated. Proofs of all these statements can be found in Section VI.1 of Analysis II by Amann and Escher.

Suppose $f\colon [a,b] \to \mathbb{K}$ is a staircase function. Let $(\xi_{0}, \ldots , \xi_{n})$ be a partition of $[a,b]$ and for each $k\in \{1, \ldots , n\}$ suppose there is $c_{k} \in \mathbb{K}$ such that $f(x) = c_{k}$ for all $x\in (\xi_{k-1}, \xi_{k})$. Define \begin{equation} \mathcal{I}(f) := \sum_{k=1}^{\ell} c_{k} (\xi_{k} - \xi_{k-1}) . \end{equation} The quantity $\mathcal{I}(f)$ is well-defined independently of the partition $(\xi_{0}, \ldots , \xi_{n})$ associated with $f$. Furthermore, $\mathcal{I}$ gives rise to a bounded linear map from the subspace of staircase functions on $[a,b]$ into $\mathbb{K}$ with norm $b-a$. The map $\mathcal{I}$ has a unique bounded extension $\tilde{\mathcal{I}}$ to the closed subspace of regulated functions on $[a,b]$ with the same norm as $\mathcal{I}$. Furthermore, the regulated functions are a proper subspace of the Riemann integrable functions on $[a,b]$, and $\tilde{\mathcal{I}}$ coincides with the Riemann integral on the subspace of regulated functions. Proofs of all these statements can be found in Sections VI.2 and VI.3 of Analysis II by Amann and Escher.

We now relate the theory of integration of regulated functions to your construction. Note that every member of $S[a,b]$ is a staircase function and that $I$ is the restriction of $\mathcal{I}$ to $S[a,b]$. It follows that every member of $\tilde{S}[a,b]$ is a regulated function and that $\tilde{I}$ is the restriction of $\tilde{\mathcal{I}}$ to $\tilde{S}[a,b]$. Hence we obtain that $\tilde{S}[a,b]$ is a proper subspace of the Riemann integrable functions on $[a,b]$ and that $\tilde{I}$ coincides with the restriction of the Riemann integral to $\tilde{S}[a,b]$. This addresses the second proposed result you were wondering about and shows that the inclusion is in fact in the opposite direction to what you originally proposed. Furthermore, by the characterisation of $\tilde{S}[a,b]$ we see that for any $c\in (a,b)$ the function $1_{\{c\}}$ is a regulated function that does not belong to $\tilde{S}[a,b]$. Hence $\tilde{S}[a,b]$ is properly contained in the space of regulated functions on $[a,b]$.

We now look at the first proposed result you were wondering about. It is true as we will show. Define the map $\Phi \colon \tilde{S}[a,b] \to L^{\infty}[a,b]$ by $\Phi (f) := f$.

We first show that $\Phi$ is injective. Let $f\colon [a,b] \to \mathbb{K}$ be left continuous at all points on $[a,b)$, right continuous at $b$ and equal to zero almost everywhere. The left continuity of $f$ on $[a,b)$ implies $f(x) = 0$ for all $x\in [a, b)$. To see this, suppose for a contradiction that $f(x_{0}) \neq 0$ for some $x_{0}\in [a,b)$. Then by right continuity at $x_{0}$ there is some $\beta \in (a,b)$ such that $f(x) \neq 0$ for all $x\in (x_{0}, \beta )$. This contradicts that $f = 0$ almost everywhere. We argue in a similar way that $f(b) = 0$. Hence we have $f(x) = 0$ for all $x\in [a,b]$. It follows that the map $\Phi$ is injective.

Moreover, if $f\in \tilde{S}[a,b]$ we have by right continuity at $b$ that $\sup_{t\in [a,b]} |f(t)| = \sup_{t\in [a,b)} |f(t)|$, and further by left continuity on $[a,b)$ that $\sup_{t\in [a,b)} |f(t)| = {\rm ess \, sup}_{t\in [a,b]} |f(t)|$. It follows that $\Phi$ is the desired isometric embedding.

Finally, here are a few additional remarks that are not directly related to your question but that might be of interest. That the functions were scalar-valued was not significant. You can replace the role of the scalars $\mathbb{K}$ with any Banach space and obtain the same result with the same proofs. This is also done in the book Analysis II by Amann and Escher. Also, note that you took the closure of $S[a,b]$ with respect to the supremum norm. If you instead took the closure with respect to the $L^{1}$-norm, you would obtain the well-known Banach space $L^{1}[a,b]$ of Lebesgue-integrable functions on $[a,b]$. See Theorem 2.26 of Real Analysis by Folland for more information about this.

In fact, Chapter X of Analysis III by Amann and Escher and Chapter VI of Real and Functional Analysis by Lang use such a procedure to develop an integration theory of functions taking values in a Banach space. Both books start off with simple functions where it is clear what the integral should be, take the completion with respect to the $L^{1}$-seminorm and then identify this completion with a suitable space of functions. The construction is at a similar difficulty to the ordinary construction of the Lebesgue integral, so I would recommend looking into this for more information if you are interested.