# Prove that the upper integral $\int_{-1}^1f(x)dx$ is equal to $0$.

Let $$f:[-1,1]\to\mathbb{R}$$ be the function defined by

$$f(x)=\begin{cases} 1&\text{ if }x=0\\ 0&\text{ if }x\neq 0.\end{cases}$$

Prove that the upper integral $$\int_{-1}^1f(x)dx$$ is equal to $$0$$.

Here is my attempt:

fix $$\epsilon > 0$$. If we compute the upper Darboux sum using the division $$D=\{-1,-\frac{\epsilon}{2},\frac{\epsilon}{2},1\}$$ we get:

$$S(D) = \sum_{i=1}^{3} \delta_i F_i$$ where $$F_i = sup\{f(x): x_{i-1} < x < x_i\}$$

$$\sum_{i=1}^{3} \delta_i F_i = (-\frac{\epsilon}{2} - (-1))F_1 + (\frac{\epsilon}{2} - (-\frac{\epsilon}{2}))F_2 + (1 -\frac{\epsilon}{2})F_3$$

since $$0 \notin[-1, -\frac{\epsilon}{2}]$$ and $$0 \notin[\frac{\epsilon}{2}, 1], F_1 = F_3= 0$$. Since $$0 \in[-\frac{\epsilon}{2}, \frac{\epsilon}{2}]$$, then $$F_2 = 1$$.

Then we have that:$$(-\frac{\epsilon}{2} - (-1))F_1 + (\frac{\epsilon}{2} - (-\frac{\epsilon}{2}))F_2 + (1 -\frac{\epsilon}{2})F_3 = (1-\frac{\epsilon}{2})(0) + (2\frac{\epsilon}{2})(1) + (1-\frac{\epsilon}{2})(0) = 0 + \epsilon$$.

It follows that $$\forall_{\epsilon>0}$$ there exists a division such that $$0 < S(D) < \epsilon$$. Also $$f(x) \geq 0$$ so $$S(D) \geq 0$$ for any division. Therefore, from the definition of the infimum $$inf\{S(D): D \text{ is a division of }[-1,1]\} = 0 = \text{ upper}\int_{-1}^1f(x)dx$$

is this proof OK?

• The general idea is correct although the notation and presentation is a bit sloppy. Mar 10, 2020 at 1:43
• That's a complicated way to say something simple. Mar 10, 2020 at 2:19
• How can I simplify it? Is it better to say that as $\epsilon$ approaches zero, the lower darboux sum approaches zero and therefore the upper integral is zero? Mar 10, 2020 at 2:33
• The lower sum is unrelated to the upper sum and upper integral. Anything you say about the lower sum means nothing about the upper sum and upper integral. Mar 10, 2020 at 2:38
• Can I say that as as ϵ approaches zero, the upper darboux sum approaches zero and therefore the upper integral is zero? Mar 10, 2020 at 2:42

You write:

It follows that $$\forall_{\epsilon>0}$$ there exists a division such that $$0 < S(D) < \epsilon$$. Also $$f(x) \geq 0$$ so $$S(D) \geq 0$$ for any division. Therefore, from the definition of the infimum $$inf\{S(D): D \text{ is a division of }[-1,1]\} = 0 = \text{ upper}\int_{-1}^1f(x)dx$$

I would write:

Thus, for all $$\epsilon > 0$$, there is a division $$D$$ such that $$S(D) = \epsilon$$. This implies that

$$\mathrm{inf} \big\{ S(D) \;\big|\; D \text{ is a partition of } [-1, 1] \big\} \leq \epsilon.$$

In the limit as $$\epsilon$$ tends to zero, we thus have

$$\mathrm{inf} \big\{ S(D) \;\big|\; D \text{ is a partition of } [-1, 1] \big\} \leq 0.$$

Since $$f(x) \geq 0$$ for all $$x$$, we have $$S(D) \geq 0$$ for any choice of division $$D$$. This means that

$$\mathrm{inf} \big\{ S(D) \;\big|\; D \text{ is a partition of } [-1, 1] \big\} \geq 0.$$

Collectively, this means the infimum is equal to zero, and so by definition of the upper integral, that is also equal to zero.