The method of proving the equality of integrals by showing they agree within $\epsilon$, for an arbitrary $\epsilon>0$

Question

Let $f,g$ be two integrable functions. To show that $\int_{a}^{b} f+g=\int_{a}^{b} f+\int_{a}^{b} g$ why do we need that for any $\epsilon>0$, $$-\epsilon+\int_{a}^{b} f+\int_{a}^{b} g<\int_{a}^{b}f+g<\int_{a}^{b} f+\int_{a}^{b} g+\epsilon\quad $$ ? Intuitively this means that their difference "goes" to $0$, but what does that mean? Is there a theorem which says two integrals $A$ and $B$ are equal if for any $\epsilon>0$, $|A-B|<\epsilon$?

Answer 1

This is true for real numbers--if A and B are real numbers and for all $\epsilon > 0$, $|A-B| < \epsilon$, then $A = B$. This is because if $A \neq B$, then $|A-B| > 0$. In particular, those integrals are real numbers, so they must be equal.

Answer 2

As explained above, the idea is that if two real numbers $a$ and $b$ are within $\epsilon$ of each other for any $\epsilon > 0$, then they are in fact equal. This is true in any metric space, since we demand that every pair of distinct elements be some positive distance apart, and then we can always take $\epsilon$ to be half of that distance.

One question you might ask is why we need to take this approach in the first place. Why can't we simply prove that $a=b$ without recourse to epsilons? The reason is that integrals themselves are defined by a limit. As such, it is hard to effectively pinpoint their value, but relatively easy to approximate it to arbitrary precision (this is built into the definition of a limit). This is why we prove equality through the limit of approximate equalities.

Answer 3

There is a theorem which says that this is true.

[Ross, Elementary analysis Ex. $3.8$] Let $a,b \in \mathbb{R}$ and $\epsilon >0$ in $\mathbb{R}$ as well.

1. If $a \le b_1$ for every $b_1>b$ then $a \le b$. This can be easily proven by contradiction.

2. Similarly if $a \ge b_1$ for every $b_1<b$ then $a \ge b$.

But if $a \le b$ and $a \ge b$ then by antisymmetry property of linearly ordered fields ($\mathbb{R}$ being one) we get that $a=b$. Apply this result with $a=\int f+g$, $b = \int f + \int g$ and $ b_1= b+ \epsilon$ for the first case and $b_1=b- \epsilon$ for the second case.

Answer 4

Let $A$ and $B$ be such that for every $\varepsilon >0$ $$|A-B|<\varepsilon$$

$$\iff A=B$$

Proof: If $A\neq B$ then choose $$\varepsilon=\dfrac{|A-B|}{2}$$ then we have $$|A-B|<\dfrac{|A-B|}{2}$$ or absurdity. A contradiction.

If $A=B$ the proof it trivial.