Why does choosing $\varepsilon$ to be arbitrarily small mean equality?

Question

Suppose that $\int_a^bf(x)dx$ exists and there is a number A such that, for every $\varepsilon> 0$ and $\delta > 0$ , there is a partition $P$ of $[a,b]$ with $||P||<\delta$ and the Riemann sum of $f$ over $P$ that satisfies the inequality $|\sigma -A|$ . Show that $\int_a^bf(x)dx=A$.

In the last part of this proof it follows that,

$|A-\int_a^bf(x)dx| \leq |A-\sigma|+|\sigma-\int_a^bf(x)dx| \leq 2\varepsilon$

Then it says we can choose $\varepsilon$ to be arbitrarily small so that $A=\int_a^bf(x)dx$.

If we choose $\varepsilon$ to be very small how does $A=\int_a^bf(x)dx$ follow?

Answer 1

Remember that $A$ and $\int_a^b f(x) dx$ are just numbers so this has nothing to do with integrals.

$$|a - b| \le \varepsilon \ \forall \varepsilon > 0 \to a=b$$

Pf:

Suppose on the contrary that $a \ne b$. We want to show that $\exists \varepsilon_0 > 0$ s.t. $|a-b| > \varepsilon_0$.

If $a \ne b$, then $|a - b| > 0$.

Choose $\varepsilon_0 = |a-b|$. ↯ QED

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To help you better understand (at least this is how I understood it when I took basic real analysis), let's see the case where $a$ and $b$ aren't just numbers.

Recall definition of $$\lim_{x \to a} f(x) = L$$

$\forall \varepsilon > 0, \exists \delta > 0$ s.t. $|f(x) - L| < \varepsilon$ wh $0 < |x-a| < \delta$

Here we don't (necessarily) have $f(x) \equiv L$ (meaning $f(x) = L \ \forall x$; hell we can have $f(x) \ne L \ \forall x$) because we don't have

'$|f(x) - L| < \varepsilon \ \forall \varepsilon > 0 \ \forall x$'

We instead have '$\forall \varepsilon > 0, |f(x) - L| < \varepsilon$ for some condition on $x$'.

Now if we have $f(x) = b$ and $\lim_{x \to a} f(x) = L$ then we have $f(x) \equiv L$ because:

$\forall \varepsilon > 0, |f(x) - L| < \varepsilon$ for some condition on $x$

$\to \forall \varepsilon > 0, |b - L| < \varepsilon$ for some condition on $x$

$\to \forall \varepsilon > 0, |b - L| < \varepsilon \ \forall x$

$\to b = L \ \forall x$

$\to f(x) = L \ \forall x$

$\to f(x) \equiv L$

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To further elaborate on $\int_a^b f(x) dx$ being just a number by putting it in the context of $\varepsilon-\delta$, you can think of

$$|A - \int_a^b f(x) dx|$$

as analogous to

$$|L - \lim_{x \to a} f(x)|$$

After all, $\int_a^b f(x) dx$ is a limit as well. As you said, $\int_a^b f(x) dx$ 'exists'.

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Finally, do you remember the proof of uniqueness of limits of functions namely

$$\lim_{x \to a} f(x) = L, \lim_{x \to a} f(x) = M$$

?

You may be interested in:

Prove that the limit of a function is unique.

Answer 2

Suppose you have a number $x \ge 0$, and I tell you that for any $ \epsilon >0$ we have $x < \epsilon$.

Then we must have $x=0$.

Suppose $x >0$ then take $\epsilon = {x \over 2}$ which gives $0 < x \le {x \over 2}$, or $1 < {1 \over 2}$ which is a contradiction.

Answer 3

If you have $|R|\leq 2\varepsilon $ for all $\varepsilon>0$, then $|R|$ is a nonnegative real number that is less than any positive real number. It follows that $R=0$. If it weren't, and $|R|>0$, we get a contradiction by taking for instance $\varepsilon=|R|/4$.

Answer 4

The left hand side is <= epsilon for ALL SMALL EPSILONS ( not just some arbitrarily small epsilon). Clearly only 0 satisfies this.

Answer 5

If $0\le |a-b| \le \epsilon $ forall $\epsilon >0$, then there is no $ \epsilon > 0$ such that $0 < \epsilon <|a-b|$. But for all $x > 0$ there does exist some $\epsilon$ so that $0 <\epsilon < x $.

So $|a-b| \not > 0 $.

So $|a-b|=0$.

Answer 6

Fundamentally this boils down to (a) the total ordering property of the real numbers, (b) the Archimedian property of the real numbers, and (c) the fact that the absolute value separates points.

Let $$x =\left|\int_a^b f(x) dx - A\right|.$$ We first observe that $x$ is a real number.

For any real number $x$, one and only one of the following properties must hold:

1. $x > 0$
2. $x = 0$
3. $x < 0$.

This follows from the fact that the real numbers are totally ordered.

We already know that $x$ is nonnegative, so that rules out the $x < 0$ (case 3.).

On the other hand, if $x>0$ (case 1.), then by the Archimedian property of the real numbers, we could find another number smaller than $x$ but still greater than zero. Whatever this number is that is supposedly smaller than $x$, we can just choose $\epsilon$ so that $2\epsilon$ is even smaller than this number. Then $x > 2\epsilon$ which contradicts the fact we already know which is that $x < 2\epsilon$. Therefore, since case (1.) leads to a contradiction, it cannot be true.

The only remaining possibility is the second case: $x=0$. Recalling the definition of $x$, this means: $$\left|\int_a^b f(x) dx - A\right|=0.$$

Now, by the fact that the absolute value function separates points, we know that for any real number $y$, if $|y|=0$ then $y=0$. Hence $$\int_a^b f(x) dx - A = 0,$$ and so $$\int_a^b f(x) dx = A.$$