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Let $f \colon\mathbb R \rightarrow \mathbb R $ be an increasing function. Define $F\colon \mathbb R \rightarrow \mathbb R $ as $$F(x)=\lim_{y \downarrow x}f(y), \text{ for every }x \in \mathbb R.$$ Prove that if $F$ is continuous then $f$ is primitivable.

My attempt:

First of all, I noticed that the value of function $F$ at an arbitrary point $x$ is equal to the right-sided limit of $f$ at the point in question.

It goes without saying that every monotonous function has a right-sided limit at every point. Therefore, I stated that the function $F$ is correctly defined.

Since the function $F$ is continuous, I guess that the function $f$ is continuous. Thus, it has to be primitivable.

But I'm not quite sure that from the continuity of $F$, we also have the continuity of $f$. May somebody help me with this part?

Sumanta
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shangq_tou
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1 Answers1

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Assume that $f$ is not continuous at some $x_0\in\mathbb R$. Then, due to the monotonicity, there is some $\varepsilon>0$ such that either $f(x_0)+\varepsilon\leq f(x)$ for all $x>x_0$, or $f(x)\leq f(x_0)-\varepsilon$ for all $x<x_0$. We only consider the first case, the second is treated similarly. Then $F(x_0)=\lim_{x\downarrow x_0}f(x)\geq f(x_0)+\varepsilon$. For $y<x<x_0$ we have $f(y)\leq f(x)\leq f(x_0)$ and hence $F(y)=\lim_{x\downarrow y}f(x)\leq f(x_0)$. Thus, $F(y)\leq f(x_0)<f(x_0)+\varepsilon\leq F(x_0)$ for all $y<x_0$, showing that $F$ is discontinuous at $x_0$.

sranthrop
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