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This question already has an answer here:

If $f$ is an increasing function over the reals, given a number $M$, is it always possible to find some $x \ge M$ such that $f$ is continuous at $x$? This seems like it should be intuitively true but I can't find a proof.

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marked as duplicate by user147263, Community Jul 12 '15 at 3:17

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the answer is affirmative. Let M>0 be given. Then f is bounded and integrable over [M,2M], being monotone. Then f is continuous almost everywhere in [M,2M] by Lebesgue's theorem. So there x>M where f is continuos.

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The following result is well known. Suppose $f$ is a real valued function monotone in the interval $[a, b], a < b$. Then $f$ can have at most only a countable number of discontinuity in $[a,b]$. Hence there are uncountably many number of points in $[a, b]$ at which $f$ is continuous.

Apply this result to an increasing function. For your question $f$ is increasing in the real numbers. So given any $M$, take the interval, say $[M, M+1]$. Then the restriction of $f$ to $[M, M+1]$ is monotone. Hence there are infinite number of points in $(M, M+1)$ at which $f$ is continuous. Of course there is an $x$ in $(M, M+1)$ ( so $x > M$) at which $f$ is continuous.

More general, if $f$ is a monotone function on the real numbers R, then $f$ is continuous almost everywhere on R, i.e., except for a subset $E$ of measure $0$, $f$ is continuous. So given $M$ , there are numbers > $M$ and numbers < $M$ and arbitrary close to $M$ at which $f$ is continuous.

For a reference to the result stated above see Theorem 3 in my article Monotone Function, Function of Bounded Variation, Fundamental Theorem of Calculus

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