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Suppose $f: \Bbb R \to \Bbb R$ is a Borel measurable function. I have to prove that $\{x: $f$ \text{ is discontinuous at } $x$\} \in B(\Bbb R)$. So, I am trying to prove that the complement event i.e. $\{x: $f$ \text{ is continuous at } $x$\}\in B(\Bbb R)$. But, $\{x: $f$ \text{ is continuous at } $x$\}=\{x: \limsup_{t->x}f(t) = \liminf_{t->x}f(t)\}$. I know that when $f$ is continuous then the functions $g(x) = \limsup_{t->x}f(t)$ and $h(x) = \liminf_{t->x}f(t)$ are measurable. But, how to tell for general $f$ ?

aaaaaa
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2 Answers2

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If you have a measurable function $f$ then the sets $\{x : f(x)>t\},\{x : f(x)=t\},\{x: f(x)<t\}$ are all measurable (as inverse images of open or closed sets). If $g,h$ are measurable then $g-h$ is measurable and $$ \{x : g(x)=h(x)\}= \{x : g(x)-h(x)=0\}$$ is therefore measurable. Apply this for $g=\limsup...$ and $h=\liminf...$ and you are done.

If $g(y)=\limsup_{t \to y} f(t)$ let's prove that $g$ is upper semicontinuous. Pick $x$ fixed and $t_n \to x$. Suppose that $g(x)<\limsup_{n \to \infty} g(t_n)$. Therefore there is a $\varepsilon>0$ and a subsequence of $(t_n)$ denoted the same for simplicity such that $g(x)+\varepsilon < g(t_n)$. For every $n$, using the definition of $g$ we can find $s_n$ as close as we want to $t_n$ such that $g(x)+\varepsilon/2<f(s_n)$. But $s_n \to x$ so taking limsup in the last inequality gives $g(x)+\varepsilon/2\leq g(x)$. Contradiction.

Therefore $g$ is upper semicontinuous and therefore measurable (Show that every upper semi-continuous real function is measurable)

The $\liminf$ $h$ is lower semicontinuous and therefore it is measurable.

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Beni Bogosel
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  • How do you say that $g$ and $h$ are measurable. I know that limsup/ liminf of a "sequence" of measurable functions is measurable. Where is sequence here ? – aaaaaa Apr 13 '14 at 14:31
  • $g$ is the liminf and $h$ is the limsup which you know are measurable. You say you need to prove that the set where $g=h$ is measurable. This is what I did above. – Beni Bogosel Apr 13 '14 at 14:34
  • As far as I know limsup of a "sequence" of measurable functions is measurable. unless we have sequence we can't say anything. Did I make my point ? – aaaaaa Apr 13 '14 at 14:44
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    Yes, I get what you mean. I don't have the time now to write the proof that if $f$ is measurable then $g(x)=\liminf_{t \to x}f(t)$ is also measurable. You can prove it using the definition of $\liminf$. If someone else doesn't do it, I will write it tonight. – Beni Bogosel Apr 13 '14 at 15:27
  • Bogosel: Thanks. – aaaaaa Apr 13 '14 at 15:37
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Define $f_{n}$ by $x\mapsto\sup\left\{ f\left(y\right)\mid y\in\left[\frac{1}{n}\lfloor nx\rfloor-\frac{1}{n},\frac{1}{n}\lfloor nx\rfloor+\frac{1}{n}\right)\right\} $. Then $\limsup f=\lim_{n\rightarrow\infty}f_{n}$. Function $f_n$ is constant on intervals $\left[\frac{k}{n},\frac{k+1}{n}\right)$ where $k\in \mathbb Z$. This because the function prescribed by $x\mapsto\lfloor nx\rfloor$ is constant on such intervals. So each $f_n$ is measurable and as a limit of measurable functions so is $\limsup f$.

drhab
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  • the sup is not over countable collections. – aaaaaa Apr 13 '14 at 16:34
  • That is not relevant here. It concerns here only the prescription of a function. – drhab Apr 13 '14 at 18:27
  • I see that Beni has helped you out, but if you have marked this answer as 'not useful' then tell me why. The argument that the sup is not over a countable union is no explanation for that. You can define a function by using such 'sups' (measurability plays no part in it yet) and secondly prove that the defined function is measurable. – drhab Apr 14 '14 at 08:17