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Suppose $f:[a,b]\to\mathbb{R}$ has a right limit $f(x+)$ at all $a \le x \lt b$ and a left limit $f(x-)$ at all $a\lt x\le b$.

Is the function $$g[a,b]\to\mathbb{R}:x\mapsto\begin{cases}f(x+)&a\le x\lt b\\ f(b)&x=b\end{cases}$$ càdlàg? and is the set $\lbrace x\in[a,b]:f(x)\neq g(x)\rbrace$ at most countable (like you would expect)?

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    There is a result which says that for any real function there exist only countably many points $x$ of $\mathbb R$ for which $f$ is not continuous at $x$ but $f(x+)$ exists. See e.g. van Rooij, Schikhof: A Second Course on Real Functions, Theorem 7.7, [p.45](http://books.google.com/books?id=Cqk5AAAAIAAJ&pg=PA45#v=onepage&q&f=false). I believe this answers the second part of your question.2012-04-26
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    Maybe I am missing something, but I think that to show that the first part is true, it suffices to show that $f(x+)=g(x+)$ and $f(x-)=g(x-)$ for each $x$, which does not seem to be that difficult.2012-04-26
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    If these two accounts [here](http://math.stackexchange.com/users/30027/martin-gale) and [here](http://math.stackexchange.com/users/26471/martin-gale) both belong to you, you might consider registering and [ask moderators](http://meta.math.stackexchange.com/questions/3542/merging-of-accounts) to [merge](http://meta.math.stackexchange.com/questions/3977/how-can-i-participate-in-the-discussion-of-my-question-if-i-lost-the-unregistere) the old accounts with the new one. Registering might make easier for you to follow the questions you posted.2012-05-01

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