# The fundamental theorem of calculus: Limit of the supremum of a function on a closed interval as the interval size approaches zero

I was reading the proof of the fundamental theorem of calculus in Spivak;

Assume we have a function $$f(x)$$ that is integrable on the closed interval $$[a,b]$$ and continuous at $$c \in [a,b]$$, suppose we have $$h>0$$, then he constructed the function:

$$M(h) = sup\{f(x): c \leq x \leq c+h\}$$

and then in the proof he said: $$\lim_{h \to 0} M(h)=f(c)$$ by the continuity of $$f$$ at $$c$$

Intuitively this is clear, but I couldn't write a proof for this limit on supremum thing.

Any clues on the proof of this limit?

Edit

From the continuity of at c, $$\lim_{x \to c} f(x) = f(c)$$, so we need to show that the limit of $$\lim_{h \to 0} M(h) = \lim_{x \to c} f(x)$$ (or $$\lim_{h \to 0} f(c+h)$$) , I couldn't establish this part

• It follows from definition of continuity at $c$. Please show your effort. Commented Nov 20, 2023 at 9:36
• from the continuity of at c, $\lim_{x \to c} f(x) = f(c)$, so we need to show that the limit of $\lim_{h \to 0} M(h) = \lim_{x \to c} f(x)$ (or $\lim_{h \to 0} f(c+h)$) , I couldn't establish this part Commented Nov 20, 2023 at 9:40
• @geetha290krm any hints? Commented Nov 20, 2023 at 10:02

First of all $$M(h)\geq f(c)$$ for any $$h$$, so $$\lim_{h\to0}M(h)\geq f(c)$$.
Then you show $$\forall\varepsilon>0$$, $$\exists H>0$$ such that $$h implies $$M(h)\leq f(c)+\varepsilon$$, which is trivial by continuity. Once you have that, letting $$h\to0$$ and $$\varepsilon\to0$$ will yield the result.