Proving that the image of a Riemann sums of a function with the intermediate value property(Darboux function) is an interval

Question

Let $a$ and $b$ be real numbers, such that $a < b$. We say $\Delta$ is a division of the interval $\left[a, b\right]$, if $\Delta=(x_0, x_1, x_2, ..., x_n)$, for some non-zero natural number $n$, where $a=x_0$, $b=x_n$, and $x_i < x_j, \forall 0 \le i < j \le n$. Informally, $\Delta=(a=x_0 < x_1 < x_2 < ... < x_n =b)$ is the set of all the end points of some intervals that partition the interval $[a, b]$.

Now let $\text{Div}(a, b)$ be the set of all divisions of the closed interval $[a, b]$. We will define $P(\Delta)$ as the set of all intermediary points for the Riemann sum. For some $\Delta \in \text{Div}(a, b)$, $P(\Delta)=\mathop{\Large\times\normalsize}_{1\le i\le n}[x_{i-1}, x_i]$. So $P(\Delta)=[x_0, x_1] \times [x_1, x_2] \times ... \times [x_{n-1}, x_n]$ is the direct product of all the partitions defined by $\Delta$.

Now for some function $f\colon [a, b] \to \mathbb{R}$, we define the Riemann sum $\sigma(f, \Delta, c)=\sum_{i=1}^n(f(c_i)(x_i-x_{i-1}))$, where $\Delta \in \text{Div}(a, b)$ and $c=(c_1, c_2, ..., c_n) \in P(\Delta)$. To make it more clear for some real valued function $f\colon[a, b] \to \mathbb{R}$ and some division of $[a, b]$ called $\Delta$, $$\small\sigma(f, \Delta, c)=f(c_1)*(x_1-x_0)+f(c_2)*(x_2-x_1)+...+f(c_i)(x_i-x_{i-1})+...+f(c_n)(x_n-x_{n-1}), \text{ where } x_{i-1}\le c_i \le x_i$$

A Darboux function, or a function that has the intermediate value property is a function defined on an interval $I$ such that $\forall x, y \in [a, b] \text{ and } \forall \lambda \text{ between $f(a)$ and $f(b)$}, \exists \xi \in [x, y] \text{ such that } f(\xi)=\lambda$.

Now for some $\Delta \in \text{Div}(a, b)$ consider $I(\Delta)=\{ \sigma(f, \Delta, c) \mid c \in P(\Delta) \}$ Show that if $f$ is a Darboux function, then $\forall \Delta \in \text{Div}(a, b), I(\Delta)$ is an interval.

My approach to this problem was to use the fact that the image through a function of a union of sets is the union of the image of each function, namely $$f(\bigcup_{i \in I}A_i)=\bigcup_{i \in I}f(A_i)$$ Now let's consider a simpler case where $\Delta=(a, x_1, b)$. Therefore, we can consider $\Delta$ and $f$ fixed and consider the function $g\colon P(\Delta) \to \mathbb{R}, \; g(c) = \sigma(f, \Delta, c)$. So $I(\Delta)=g(P(\Delta))$ $$P(\Delta)=\bigcup_{\beta \in [x_1, b]}\left( \bigcup_{\alpha \in [a, x_1]} \{ \alpha \} \times \{ \beta \} \right) \implies$$ $$g(P(\Delta))=g\left(\bigcup_{\beta \in [x_1, b]}\left( \bigcup_{\alpha \in [a, x_1]} \{ \alpha \} \times \{ \beta \} \right)\right) = \bigcup_{\beta \in [x_1, b]}\left( \bigcup_{\alpha \in [a, x_1]} g(\{ \alpha \} \times \{ \beta \}) \right)=\bigcup_{\beta \in [x_1, b]}\left( \bigcup_{\alpha \in [a, x_1]} \{ f(\alpha)(x_1-a)+f(\beta)(b-x_1) \} \right)$$ Let's evaluate $T=\bigcup_{\alpha \in [a, x_1]} \{ f(\alpha)(x_1-a)+f(\beta)(b-x_1) \}$.

Claim 1:

Case1: If there is $u, v \in [a, x_1]$ such that $f(u)=\inf_{x\in [a, x_1]} f(x)$ and $f(v)=\sup_{x\in [a, x_1]} f(x)$, then $T$ is a closed interval and $$T=[\inf_{x\in [a, x_1]}f(x)(x_1-a)+f(\beta)(b-x_1), \sup_{x\in [a, x_1]}f(x)(x_1-a)+f(\beta)(b-x_1)\big]$$

Case2: If there is $u \in [a, x_1]$ such that $f(u)=\inf_{x\in [a, x_1]} f(x)$, but no $v$ satisfying the condition above, then $T$ is a semiclosed interval and $$T=\big[\inf_{x\in [a, x_1]}f(x)(x_1-a)+f(\beta)(b-x_1), \sup_{x\in [a, x_1]}f(x)(x_1-a)+f(\beta)(b-x_1)\big)$$

Case3: If there is $v \in [a, x_1]$ such that $f(v)=\sup_{x\in [a, x_1]} f(x)$, but no $u$ satisfying the condition above, then $T$ is a semiclosed interval and $$T=\big(\inf_{x\in [a, x_1]}f(x)(x_1-a)+f(\beta)(b-x_1), \sup_{x\in [a, x_1]}f(x)(x_1-a)+f(\beta)(b-x_1)\big]$$

Case4: If there are no such $u, v$ then $T$ is an open interval and $$T=\big(\inf_{x\in [a, x_1]}f(x)(x_1-a)+f(\beta)(b-x_1), \sup_{x\in [a, x_1]}f(x)(x_1-a)+f(\beta)(b-x_1)\big)$$.

In the claim above I am considering the cases where $\inf f(x)=-\infty$ or $\sup f(x) = +\infty$.

As for the proof:

Let $U$ be the be the set, whose elements are all the real numbers between $$\inf_{x\in [a, x_1]}f(x)(x_1-a)+f(\beta)(b-x_1) \text { and }\sup_{x\in [a, x_1]}f(x)(x_1-a)+f(\beta)(b-x_1)$$ I am disregarding, for the moment, whether it is a closed interval or not(so the proof will be easier to follow). I will also denote $m=\inf_{x \in [a, x_1]} f(x)$ and $M=\sup_{x \in [a, x_1]} f(x)$. For any of the four cases, it is clear that $T \subseteq U$.

Using the fact that $f$ is a Darboux function I will prove $U \subseteq T$. Let's assume we are in Case 1. Therefore we are considering $U$ as a closed interval: $$\text{Let }h \in U \implies m(x_1-a)+f(\beta)(b-x_1) \le h \le M(x_1-a)+f(\beta)(b-x_1) \implies$$ $$f(u) \le \frac{h-f(\beta)(b-x_1)}{x_1-a} \le f(v) \implies \exists w \in [u, v] \text{ such that } f(w)=\frac{h-f(\beta)(b-x_1)}{x_1-a} \implies h=f(w)(x_1-a)+f(\beta)(b-x_1) \implies h \in T \implies U \subseteq T$$ The other cases seem intuitive, but I can't seem to deduce a rigurous proof: For example let's we assume $m$ is finite and we are in Case 3. We can follow the same steps as in Case1, if we rigorously conclude the following statement, $$m < \frac{h-f(\beta)(b-x_1)}{x_1-a} \leq M=f(v) \implies \exists w \in [a, x_1] \text{ such that } f(w)=\frac{h-f(\beta)(b-x_1)}{x_1-a}$$

The proof for the other cases should be similar.

Claim 2:

$\bigcup_{\beta \in [x_1, b]}T \text{ is an interval } \text{whose lower bound is } $ $$\inf_{x \in [a, x_1]}f(x)(x_1-a)+\inf_{x \in [x_1, b]}f(x)(b-x_1)$$ $\text{ and whose upper bound is }$ $$\sup_{x \in [a, x_1]}f(x)(x_1-a)+\sup_{x \in [x_1, b]}f(x)(b-x_1)$$ The interval metioned above is closed at its lower bound iff $$\exists p \in [a, x_1] \text{ and } \exists q \in [x_1, b] \text{ such that }f(p)=\inf_{x \in [a, x_1]}f(x) \text{ and }f(q)=\inf_{x \in [x_1, b]}f(x)$$ The interval is closed at its upper bound iff $$\exists p \in [a, x_1] \text{ and } \exists q \in [x_1, b] \text{ such that }f(p)=\sup_{x \in [a, x_1]}f(x) \text{ and }f(q)=\sup_{x \in [x_1, b]}f(x)$$

This seems intuitive, but how would I go about proving this. And if we somehow manage to show that what I said is true, then can we do induction on the number of elements of some $\Delta \in \text{Div}(a, b)$ and then draw the conclusion for all elements of $\text{Div}(a, b)$?

Any tips or solutions will be gladly welcomed. Thank you for your time.

Answer 1

To show that if f is a Darboux function, then $I(Δ)$ is an interval for any $Δ∈Div(a,b)$, we need to show that I(Δ) satisfies the definition of an interval.

First, note that since Δ is a division of [a,b], we have $a = x_0 < x_1 < ... < x_n = b$, and the sets $[x_{i-1},x_i]$ for $i=1,2,...,n$ form a partition of $[a,b]$. Thus, $P(Δ) = [x_0,x_1] × [x_1,x_2] × ... × [x_{n-1},x_n]$.

Now, let $I = I(Δ)$ be the set of values of $σ(f,Δ,c)$ as c varies over $P(Δ)$. To show that I is an interval, we need to show that for any x, y in I with x < y, and any z such that x < z < y, we have z in I as well.

So let $x = σ(f,Δ,c_1)$ and $y = σ(f,Δ,c_2)$ be two elements of $I$ with $x < y$, and let z be any number such that $x < z < y$. We need to show that z is in $I$ as well, i.e., there exists a point $c_3$ in $P(Δ)$ such that $σ(f,Δ,c_3) = z$.

Since f is a Darboux function, we know that for any λ between f(a) and f(b), there exists a point ξ in [a,b] such that $f(ξ) = λ$. In particular, since x and y are in $I$, there exist points $c_1'$ and $c_2'$ in $P(Δ)$ such that $σ(f,Δ,c_1') = x$ and $σ(f,Δ,c_2') = y$.

Now consider the function $g(t) = σ(f,Δ,c)$ for $c = (c_1',c_2',...,c_{n-1}',t)$, i.e., g(t) is the Riemann sum of f over the partition $[x_0,x_1], [x_1,x_2], ..., [x_{n-1},t], [t,x_n]$, where the points $c_1', c_2', ..., c_{n-1}'$ are fixed, and only the last point t varies.

Note that $g(x_0) = σ(f,Δ,c_1') = x$ and $g(x_n) = σ(f,Δ,c_2') = y$. Also, since f is a Darboux function, g(t) is a continuous function of t. Thus, by the intermediate value theorem, there exists a point $c_3$ in $[x_0,x_n]$ such that $g(c_3) = z$.

But this means that $σ(f,Δ,c_3) = z$, where $c_3 = (c_1',c_2',...,c_{n-1}',t)$ for this particular value of t. Therefore, z is in $I$ as well, and we have shown that $I$ is an interval.

Answer 2

Let us simplify the notation a bit. A division or partition of $[a, b] $ is a finite set of points from $[a, b] $ which necessarily includes the end points $a, b$. We usually write a partition $P$ of $[a, b] $ as $$P=\{x_0,x_1,x_2,\dots,x_n\} $$ where $$a=x_0<x_1<x_2<\dots<x_n=b$$ The partition $P$ divides the interval $[a, b] $ into $n$ subintervals $[x_{k-1},x_k]$ for $k=1,2,\dots,n$. The advantage of set notation over tuple notation is that it allows union and intersection of partitions.

For a given partition $P$ we introduce a set $$C_P=\{c_1,c_2,\dots, c_n\} $$ of tags $c_k$ where $c_k\in[x_{k-1},x_k]$. Given a function $f:[a, b] \to\mathbb{R} $ we define a Riemann sum $$S(f, P, C_P) =\sum_{k=1}^n f(c_k) (x_k-x_{k-1})$$ When $f, P, C_P$ are available from context we can just use $S$ instead of $S(f, P, C_P) $.

Let us note that if $c\in(a, b) $ (ie $c$ lies in interior of $[a, b]$) and $P_1,P_2$ are partitions of $[a, c], [c, b] $ respectively then $P=P_1\cup P_2$ is a partition of $[a, b] $. And if $S(f, P_1,C_{P_1}),S(f,P_2,C_{P_2})$ are any Riemann sums of $f$ over $[a, c], [c, b] $ respectively then $$S(f, P, C_P) =(f, P_1,C_{P_1}) +S(f, P_2,C_{P_2})$$ is a Riemann sum of $f$ over $[a, b] $ where $C_P=C_{P_1}\cup C_{P_2}$. Note further that we can split a given Riemann sum $S(f, P, C_P) $ in above manner by splitting the interval $[a, b] $ into $[a, x_k], [x_k, b] $ with $c=x_k$.

Let us now assume that $f$ is a Darboux function ie it satisfies the intermediate value property. If we consider partition with just 1 subinterval then a Riemann sum is of the form $f(c) (b-a) $ and the values of these sums form an interval because of the intermediate value property of $f$. Let us now use induction on number $n$ of subintervals created by partition $P$ and assume that the result in question holds for all partitions with less than $n$ subinterval.

Let $P$ be a partition of $[a, b] $ with $n$ subintervals $[x_{k-1},x_k]$ and let $S=S(f, P, C_P), S'=f(f, P, C'_P) $ be two Riemann sums based on two tags $C_P, C'_P$. Let us choose a specific $k$ with $1\leq k<n$ and split the interval $[a, b] $ into two as $[a, x_k], [x_k, b] $. This allows us to write $P=P_1\cup P_2$ where $P_1,P_2$ are partitions of $[a, x_k], [x_k, b] $ respectively and $P_1,P_2$ induce $k, n-k$ subintervals respectively. A similar split is available for the tag sets $C_P, C'_P$ as well and let us write $$S_1=S(f,P_1,C_{P_1}),S_2=S(f,P_2,C_{P_2}),S'_1=S(f,P_1,C'_{P_1}),S'_2=S(f,P_2,C'_{P_2})$$ so that $$S=S_1+S_2,S'=S'_1+S'_2$$ Any number between $S, S'$ can be written as $tS+(1-t)S'$ where $t\in[0,1]$ and this equals $$[tS_1+(1-t)S'_1] +[tS_2+(1-t)S'_2] $$ Now $tS_1+(1-t)S'_1$ is a number lying between two Riemann sums $S_1,S'_1$ for $f$ for partition $P_1$ (with $k$ subintervals) over $[a, x_k] $. By induction hypotheses this can be written as a Riemann sum over $P_1$. Similarly $tS_2+(1-t)S'_2 $ can be written as a Riemann sum over $P_2$. And therefore their sum which equals $tS+(1-t)S'$ can be written as a Riemann sum over $P$. It follows by induction that the set of values of Riemann sums over $P$ is an interval.