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Let $f$ be continous at $c$. Prove $$\lim_{h \to 0} \left(\inf \,\{f(x)\mid c \leqslant x \leqslant c+h\}\right)=f(c)$$

This fact is used in Spivak's book to prove 1nd Fundamental Calculus Theorem.

This question is duplicated, I have done it previously but i didn't understand yet

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The inf is bounded from below for every $h > 0$ because f is continous (can you show why?). Assuming $$ \lim_{h \rightarrow \infty} \inf \{ f(x) : c \leq x \leq c + h \} =: K \neq f(c), $$ then there exits some sequence $(x_n)_n$ such that for all $n \in \mathbb{N} : x_n > c, \lim_{n \rightarrow \infty} x_n = c$ and $\lim_{n \rightarrow \infty} f(x_n) = K \neq f(c) = f(\lim_{n \rightarrow \infty} x_n)$ a contradiction to the continuity of $f$.

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Actually, that's not the only thing Spivak uses in his proof. He first considers the case $h > 0$, then $h < 0$, and then claims that $\lim \limits_{h \to 0} m_h = 0$. So, to take into account both signs of $h$, lets do the following: suppose $a < c < b$, and for any $h \in \Bbb{R}$ , define the set \begin{equation} A_h = \{f(x): a \leq x \leq b \quad \text{and} \quad |x-c| \leq |h| \}. \end{equation} First, since $f$ is integrable, it is by definition bounded, so for all $h \in \Bbb{R}$, $A_h$ is a non-empty, bounded set hence $m_h := \inf A_h$ exists. We now wish to use continuity of $f$ at $c$ to prove $\lim \limits_{h \to 0}m_h = f(c)$.

To do this, let $\varepsilon > 0$ be arbitrary. By assumption, $f$ is continuous at $c$, so there is a $\delta > 0$ such that for all $x \in [a,b]$, if $|x-c| < \delta$, then $|f(x) - f(c)| < \varepsilon$. Now, let $0 < |h| < \delta$ be arbitrary. Then, for every $x \in [a,b]$ which satisfies $|x-c| \leq |h|$, we have that (since $|h| < \delta$) \begin{equation} f(c) - \varepsilon < f(x) < f(c) + \varepsilon. \end{equation} What this says is that $f(c) - \varepsilon$ is a lower bound for $A_h$ and $f(c) + \varepsilon$ is an upper bound for $A_h$. Hence, it follows by definition of infimium that \begin{equation} f(c) - \varepsilon \leq m_h \leq f(c) + \varepsilon \end{equation} Or equivalently, $|m_h - f(c)| \leq \varepsilon$. This completes the proof, because we have shown that given any $\varepsilon > 0$, there is a $\delta > 0$ such that for all $h \in \mathbb{R}$, if $0 < |h| < \delta$, then $|m_h - f(c)|\leq \varepsilon$.


Note: From my proof, we can only conclude that $|m_h - f(c)| \leq \varepsilon$, not $|m_h - f(c)| < \varepsilon$. But this still shows the desired limit because this is true for every $\varepsilon > 0$. If this isn't clear to you, see one of the exercises in Chapter 5 of Spivak (it's probably 25 or 26).

Next, for $c=a$ or $c=b$ you only have to modify the proof slightly; and for the proof of $\lim \limits_{h \to 0} M_h = f(c)$, you just have to replace $m_h$ with $M_h$ everywhere above, and the same proof works.

Also, while it's important to be able to prove this, I remember when I read his proof of the FTC, this confused me immensely. I think there is a simpler, more direct proof though, and I can outline it if you wish.

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