If $f(x) \leq a \leq g(x)$, $f(x) \leq b \leq g(x)$, and for all $\epsilon > 0$ there exists $x$ such that $g(x) - f(x) < \epsilon$, then $a = b$? Suppose $f, g$ are real-valued functions and $a, b$ are real numbers. If $f(x) \leq a \leq g(x)$ and $f(x) \leq b \leq g(x)$ for every $x \in \mathbb{R}$, and for all $\epsilon > 0$ there exists $x$ such that $g(x) - f(x) < \epsilon$, then $a = b$.
Is this result true? Something similar to it but involving lower and upper sums came up in a proof for the sum rule in integration ($\int_{a}^{b} (f + g) = \int_{a}^{b} f + \int_{a}^{b} g$) in Spivak's Calculus.
 A: The result is true. Let $\epsilon > 0$, and choose $x \in \mathbb{R}$ such that:
$$
g(x) - f(x) < \epsilon
$$
Remark that from your conditions, $g(x) \geq a$ and $-f(x) \geq -b$, so we have:
$$
\epsilon > g(x) - f(x) \geq a-b
$$
Similarly, we have $g(x) \geq b$ and $-f(x) \geq -a$, so:
$$
\epsilon > g(x) - f(x) \geq b-a = -(a-b)
$$
And so we have:
$$
|a-b| < \epsilon
$$
Since this holds for any $\epsilon > 0$, we can conclude that $a=b$.
A: Note: $|b-a| \le g(x) - f(x)$ for all $x$.  So for all $\epsilon$ there is an $x$ so that $\epsilon > g(x) - f(x) \ge |b-a|$.   So $0 \le |b-a| < \epsilon$ for all positive $\epsilon$.  There is only one non-negative real number that is smaller than all positive numbers.
A: Without loss of generality assume that $a\leq b$. Then you have
$$
f(x)\leq a \leq b \leq g(x) \quad \text{for every}  \ x,
$$
and therefore
$$
0\leq b-a\leq g(x)-f(x) \quad \text{for every} \ x,
$$
If for every $\varepsilon >0$ there exists $x$ such that $g(x)-f(x)< \varepsilon$, we conclude that
$$
0\leq b-a \leq \varepsilon \quad \text{for every} \ \varepsilon >0,
$$
and then $b=a$.
A: Given your scenario, we can find a sequence $x_n$ so that $\lim_{n\to\infty}|f(x_n)-g(x_n)| =0$.
Therefore, $a=b$ holds.
