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I have the following exercise from the book "A course on real analysis":

Show that $\chi_I$ is Borel measurable for each open interval $I$, where $\chi_I$ denotes the indicator function of $I$.

Now, the way a Borel measurable function is defined in this book, is that it is a memeber of $\hat C$, the smallest collection of functions on $\mathbb{R}$ that contains the collection of continous functions and is closed under pointwise limits.

Since $\chi_I$ is obviously not continous on $\mathbb{R}$ in general, I have tried to show that it is a limit of a series of functions by constructing such a series explicitly, but without any luck. Maybe there is an even simpler way?

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Let $I=(a,b)$ and $d=\frac{b-a}{3}>0$. For $n\geq 1$, let $a_n=a+\frac{d}{n}$ and $b_n=b-\frac{d}{n}$ and consider the function $f_n$ defined as follows:

  • $f_n(x)=0$ for $x\leq a$ and $x\geq b$,
  • $f_n(x)=1$ for $a_n\leq x\leq b_n$,
  • $f_n$ is linear for $a\leq x\leq a_n$ and $b_n\leq x\leq b$.

Then $f_n$ is continuous and converges pointwise to $1_I$. If you draw a picture, then this construction is quite obvious.


p.s. An alternative definition of Borel measurability for $f:\mathbb{R}\to\mathbb{R}$ is is that $f^{-1}(A)$ is Borel-measurable whenever $A$ is Borel-measurable. But if $f=1_I$ then depending on whether $A$ includes $1$, $0$, both, or neither, $f^{-1}(A)$ can only be $I$, $\mathbb{R}-I$, $\mathbb{R}$, and $\emptyset$. All four possibilities are Borel-measurable, so $1_I$ is Borel-measurable.

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  • 2
    $\begingroup$ Awesome. Didn't know that the great leader is good at math as well! $\endgroup$ – Slug Pue Sep 2 '14 at 23:29
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Let $(a,b)$ be a non-empty interval, and let $0 < \epsilon < (b-a)/2$. Define a piecewise linear function $f_{\epsilon}$ to be 0 for $x \le x$ or $x \ge b$, and $1$ in $(a+\epsilon,b-\epsilon)$. Let $\epsilon \downarrow 0$.

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