Why doesn't the upper integral of a stirctly positive function $f$ tend to $0$?

Question

The definition of upper integral for a bounded function $f$ on $[a,b]$ states that

$$\overline{\int_a^b} f(x) dx = \inf \{ U(f, P): P \text{ is a partition of } [a,b] \}$$

where $U(f,P)$ is the upper sum of $f$ with respect to parition $P$.

Now suppose that $f$ is a strictly positive function on $[a,b]$ (say $f > 1 \forall x \in [a,b])$. How do we know that $\inf \{ U(f, P): P \text{ is a partition of } [a,b] \} \neq 0$? Since as we take a more refined $P$, the length of subintervals $[x_{i-1},x_i]$ get smaller and smaller. That is $\Delta x_i = x_i - x_{i-1} \rightarrow 0$ as $P$ is getting more refined. I understand that just because $\Delta x_i \rightarrow 0$, it would not necessarily imply that $\Delta x_i \times \sup \{f(x): x \in [x_{i-1}, x_i]\} \rightarrow 0$.

Can someone please provide a formal argument as to why $\Delta x_i \times \sup \{f(x): x \in [x_{i-1}, x_i]\} \not\rightarrow 0$?

I understand similar question can be posed for the lower integral, but for the sake of being specific, I hope the answer to this question clarifies my similar questions.

Answer 1

It's probably not going to be possible to provide a proof for an arbitrary function $f$ because it's possible to define pathological functions that aren't even Riemann integrable, or which take arbitrarily small values so that the upper Riemann sum does actually tend to zero.

However, here is a formal proof for a particular set of functions - namely, the set of functions that are (a) Riemann integrable on the interval $[a, b]$ (so we know that the set of upper sums has a well-defined infimum) and (b) take a well-defined minimum value on the interval $[a, b]$, i.e. there is a point $x_0 \in [a, b]$ and a value $L$ such that $f(x_0) = L$ and $x \in [a, b] \implies f(x) \geq L$. Note that since $f$ is a strictly positive function, $L > 0$.

Then, for any partition $P$ of the interval, we can say:

$$\begin{eqnarray} U(f, P) & = & \sum f(x_i^*) \Delta {x_i} \\ & \geq & \sum L \Delta x_i \\ & = & L \sum \Delta x_i \\ & = & L \left[(x_1 - x_0) + (x_2 - x_1) + \ldots + (x_n - x_{n-1}) \right] \\ & = & L (x_n - x_0) \\ & = & L (b - a) \\ \end{eqnarray}$$

In the second line, we use the fact that $f(x_i^*)$ (the largest value of $f(x)$ in the $i$th interval) must be at least $L$, since that's the lower bound for the function.

So for this function, every upper sum is bounded from below by $L(b-a)$, so the infimum of the set of all upper sums is also bounded by this value, which is strictly greater than zero.

Answer 2

I understand that just because $\Delta x_i \rightarrow 0$, it would not necessarily imply that $\Delta x_i \times \sup \{f(x): x \in [x_{i-1}, x_i]\} \rightarrow 0$.

That understanding is incorrect. If the function $f$ is bounded on $[a,b]$, and if $P$ is successively refined so that $\Delta x_i \to 0$, where $\Delta x_i = x_i - x_{i-1}$ for each partition $P$, then $\Delta x_i \times \sup \{f(x): x \in [x_{i-1}, x_i]\} \rightarrow 0$.

But the unstated yet fatal assumption in your question came out only in a comment:

but $\Delta x_i \rightarrow 0$, so how do we know that $\Delta x_i \cdot 1 \rightarrow 0$ for the choice of $P$ that gives us the infimum?

The fatal words are "the choice of $P$ that gives us the infimum". In the general case, there is no such thing.

Every choice of $P$ is a partition of $[a,b]$ into a finite number of sub-intervals leading to the approximation of the upper integral by the sum of areas of a finite number of rectangles. If (for example) $f > 1$ on $[a,b]$ and you have a partition $P$ with $n$ sub-intervals, we end up with \begin{align} U(f, P) &= \sum_{i=1}^n (\Delta x_i) \sup \{f(x): x \in [x_{i-1}, x_i]\}\\ &\geq \sum_{i=1}^n \Delta x_i \\ &= (x_1 - x_0) + (x_2 - x_1) + \cdots + (x_n - x_{n-1}) \\ &= (x_n - x_{n-1}) + \cdots + (x_2 - x_1) + (x_1 - x_0) \\ &= x_n - x_0 \\ &= b - a \end{align}

or, in short, $U(f, P) \geq b - a$, which is a positive number, for every partition $P$. Therefore the infimum of $U(f, P)$ cannot be zero.

But even for as simple a function as $f(x) = 1 + 2x$ integrated on the interval $[a,b] = [0,1]$, for which the infimum of $U(f, P)$ happens to be $2$, no partition $P$ can give us a set of upper rectangles whose total area is exactly equal to $2$. That's not what we mean when we say $2$ is the infimum of $U(f, P)$.

The infimum exists in the example $f(x) = 1 + 2x$ not because there is any partition $P$ for which $U(f, P) = 2$, but because for any partition $P$ for which $U(f, P) = 2 + \varepsilon$, no matter how small $\varepsilon$ is, there is another partition $P'$ for which $U(f, P') < U(f, P)$, thereby demonstrating that $2 + \varepsilon$ is not the infimum. But also there is no partition $P$ such that $U(f, P) < 2$, for if there were such a partition then $2$ would not be the infimum.

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Naturally, that is not a formal proof that the upper integral of $f(x) = 1 + 2x$ on $[0,1]$ is equal to $2$. But now that you know what that statement does not say, you may have a chance to demonstrate to yourself that what the statement does say is true.

And in general we do not need to be concerned about proofs where you assume any arbitrary positive function $f$ and then start asking questions about $\inf \{ U(f, P): P \text{ is a partition of } [a,b] \}$. The fact is that there are plenty of functions that are non-integrable, and finding $\overline{\int_a^b} f(x)\,\mathrm dx$ is not much use for those functions. What is much more important is that there are many functions for which we can compute both $\overline{\int_a^b} f(x)\,\mathrm dx$ and $\underline{\int_a^b} f(x)\,\mathrm dx$ and confirm that they are equal, and then we have the integral of a function. There are also classes of functions (for example, continuous bounded functions on a closed interval) for which $\overline{\int_a^b} f(x)\,\mathrm dx$ always exists, although proving that may be something for a textbook rather than a stackexchange answer.