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Take $f: (a,b) \to \mathbb{R}$ , continuous for all $x_{0}\in (a,b)$ and take $(Ω = (a,b) , F = ( (a,b) ⋂ B(\mathbb{R}))$ where $B(\mathbb{R})$ is the Borel $\sigma$-algebra.

Prove $f$ is a borel function by showing that $\{x \in(a,b): f(x) < c \}$ is in $F$.

I know that continuity of f means that for all $x\in(a,b)$ and all $\varepsilon>0$ there exists a $\delta>0$ such that $|x-x_{0}| < \delta$ implies $|f(x)-f(x_{0})| < \varepsilon$.

But Then I am stuck, how would I use these facts to help me ?

Thanks in advance for any help

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3  
Hint: You may want to use the fact that the pre-image of an open set is open under a continuous function. –  Thomas E. May 20 '12 at 19:28
    
Are you familiar with the facts that (1) the preimage of an open set (under a continuous function) is open and that (2) open subsets of the reals are elements of the Borel $\sigma$-algebra? If so, that should be all you need in order to accomplish your task. If you aren't familiar with the first fact, it would be a good exercise to show that the "$\delta,\varepsilon$" definition of continuity is in fact equivalent to the "preimages of open sets are open" definition. –  Cameron Buie May 20 '12 at 19:35

1 Answer 1

up vote 3 down vote accepted

To expand on Thomas E.'s comment: if $f$ is continuous, $f^{-1}(O)$ for $O$ open is again open. $\{x \in (a,b) : f(x) < c \} = f^{-1}((- \infty , c)) \cap (a,b)$. Now all you need to show to finish this proof is that $f^{-1}((- \infty , c))$ is in the Borel sigma algebra of $\mathbb R$.

Edit (in response to comment)

Reading your comment I think that your lecturer shows that $S := \{x \in (a,b) : f(x) < c \} $ is open. In a metric space, such as $\mathbb R$ with the Euclidean metric, a set $S$ is open if for all $x_0$ in $S$ you can find a $\delta > 0$ such that $(x_0-\delta, x_0+\delta) \subset S$.

To show this, your lecturer picks an arbitrary $x_0 \in S$. Then by the definition of $S$ you know that $f(x_0) < c$. This means there exists an $\varepsilon > 0$ such that $f(x_0) + \varepsilon < c$, for $\varepsilon$ small enough. Since $f$ is continuous you know you can find a $\delta_1 > 0$ such that $x \in (x_0 - \delta_1, x_0 + \delta_1) $ implies that $|f(x_0) - f(x)| < \varepsilon$. Now you don't know whether $(x_0 - \delta_1, x_0 + \delta_1) $ is contained in $(a,b)$. But you know that since $(a,b)$ is open you can find a $\delta_2 > 0$ such that $(x_0 - \delta_2, x_0 + \delta_2) \subset (a,b)$. Now picking $\delta := \min (\delta_1, \delta_2)$ gives you that $(x_0 - \delta, x_0 + \delta) \subset S$ because $(f(x_0) - \varepsilon, f(x_0) + \varepsilon) \subset (-\infty , c)$.

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If you can't finish the proof ping me and I'll add the rest. –  Matt N. May 20 '12 at 19:36
    
Okay, the only thing is, my lecturer has given a rough proof which we are meant to follow, but I dont understand it! but it follows the route of using the epsilon delta definition of continuity and also the definition of open sets which involes deltas too. Do you have any idea how I could prove it like this? Thanks very much for your replies! –  Rosie May 20 '12 at 20:15
    
@Rosie An equivalent way of saying $f$ is continuous on $\mathbb R$ is to say that for every $\varepsilon > 0$ there is a $\delta > 0$ such that $f(B(x_0, \delta)) \subset B(f(x_0), \varepsilon)$. Is this what you mean? –  Matt N. May 20 '12 at 20:28
    
Sorry, Im not being very clear I know, well he basically gives us the continuity definition I wrote above, and then he says choose x0 in { f < c } , this means that 0 < c-f(x0) = 2ε f is continuous so there exists δ1>0 such that for all ε>0 and for all x in (a,b) |f(x)-f(x0)| < ε, (a.b) open implies there exists δ2>0 such that (x0-δ2,x0+δ2) is a subset of (a,b) Take δ = min{δ1.δ2}then this implies (x0-δ,x0+δ)is a subset of { f < c } . I just can't follow it! –  Rosie May 20 '12 at 20:38
    
@Rosie I think there are some things that aren't clear in your lecturer's proof. Have a look at what I added in the answer and let me know if it clears up things. –  Matt N. May 20 '12 at 21:01

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