# How do you modify the value of the function on set of zero measure ( $[0,\frac 12]$) such that it become continuous function on $[0,1]$?

$$f(x) = \begin{cases} 1 & x \in [0,\frac 12] \\ 0 & x \in (\frac 12, 1] \end{cases}$$

The function is discontinuous at $\frac 12$. How do you modify the value of this function on set of zero measure ( $[0,\frac 12]$) such that we get continuous function on $[0,1]$?

I try to modify this function, but I cannot change it become continuous function. So I think it cannot modify the value on set of zero measure to become the continuous function. Could you help me? Thank you so much.

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It's not possible. – user61527 Jan 1 '14 at 21:39
Why do you think this is possible? (It's not.) – Potato Jan 1 '14 at 21:39
It is impossible. No matter how you change $f$ on a set of measure zero, it will still be discontinuous at $x={1 \over 2}$. – copper.hat Jan 1 '14 at 21:39
You can modify the function of a set of arbitrarily small measure and make it continuous. – copper.hat Jan 1 '14 at 21:40
Could you please show me more detail with the above function? – user52523 Jan 1 '14 at 21:51

The goal isn't possible. Suppose that we could do this; since $f$ is $1$ almost everywhere on $[0, \frac 1 2]$, we can find a sequence $x_n \to \frac 1 2$ from below such that $f(x_n) = 1$ for all $n$. Likewise find a sequence $y_n \to \frac 1 2$ from above for which $f(y_n) = 0$ for every $n$. Hence the function is not continuous at $\frac 1 2$.
You can change the function on a set of arbitrarily small measure by inserting a little line segment connecting the points $(\frac 1 2 - \epsilon, 1)$ and $(\frac 1 2 + \epsilon, 0)$.
@user52523 Changing on a set of measure $0$ does very little to the function. There are points not changed on every interval contained in $[0, 1]$. – user61527 Jan 1 '14 at 22:00
@user52523 Replace the function on the interval $(1/2 - \epsilon, 1/2 + \epsilon)$ with the line segment; you'll end up with a function which is piecewise linear. As far as related references, any book on basic measure theory would be good, such as Folland, Rudin or Royden. – user61527 Jan 1 '14 at 22:14