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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?

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  • $\begingroup$ The general idea is correct although the notation and presentation is a bit sloppy. $\endgroup$
    – Math1000
    Mar 10, 2020 at 1:43
  • $\begingroup$ That's a complicated way to say something simple. $\endgroup$
    – zugzug
    Mar 10, 2020 at 2:19
  • $\begingroup$ 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? $\endgroup$ Mar 10, 2020 at 2:33
  • $\begingroup$ 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. $\endgroup$
    – Jivan Pal
    Mar 10, 2020 at 2:38
  • $\begingroup$ Can I say that as as ϵ approaches zero, the upper darboux sum approaches zero and therefore the upper integral is zero? $\endgroup$ Mar 10, 2020 at 2:42

1 Answer 1

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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.

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