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Suppose that we choose some interval and fix it, for example let us choose interval $[0,1]$.

If $f$ is some everywhere continuous and nowhere differentiable function defined on $[0,1]$, then, because it is continuous the Riemann integral of it exists.

I would like to know the following:

Is it true that for every $\alpha \in \mathbb R$ there exists everywhere continuous and nowhere differentiable function defined on $[0,1]$ such that we have $\int_{0}^{1}f(x)dx=\alpha$ and $f(0)=0$ and $f(1)=1$?

I do not know how to prove this in some nice way, maybe we could argue that if we choose some everywhere continuous and nowhere differentiable function $g$ defined on $[0,1]$ and denote its integral as $\alpha^*$ then we could deform $g$ continuously to obtain some other functions which have the integral equal to any real number we want.

And even if my stream of reasoning is good I do not know how to formalize it appropriately.

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  • $\begingroup$ My idea would be something like this: let $f(x)$ satisfy for $\alpha = 1$, then for other $\alpha$ take $g(x)$ to be the limit of $g_n(x)$ where $g_n(x) = \alpha f(x)$ on $[0, 1 - \frac{1}{n}]$ and $(\alpha n (1 - x) + n(x - (1 - \frac{1}{n}))f(x)$ on $[1 - \frac{1}{n}, 1]$. Not sure if this limit function would also be nowhere differentiable, though. $\endgroup$
    – MCT
    May 16 '16 at 15:54
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Suppose you have a continuous nowhere differentiable $f$, then if $g$ is differentiable, we must have that $f+g$ is continuous nowhere differentiable.

Then for any $a,b,c$ we can find a second degree polynomial $g$ such that $g(0)=a, g(1) = b$ and $\int g = c$ (infact, with a little effort we can produce an explicit formula for the coefficients of $g$ in terms of $a,b,c$).

Since $\int (f+g) = \int f +c$, $(f+g)(0) = f(0)+a$, $(f+g)(1) = f(1)+b$, we can choose $g$ so that these have any prescribed values.

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Let $f$ be some everywhere continuous nowhere differentiable function such that $\int_0^1 f = \beta$ and consider $g$ a polynomial given by $g(x)=ax^2+bx+c$.

The function $h=f+g$ will be continuous and also nowhere differentiable, for if it were differentiable at a point, then the difference $h-g=f$ would also be at that point. We would like $h$ to satisfy:

  • $h(0)=0$, that is, $g(0)= c =-f(0)$
  • $h(1)=0$, that is, $g(1)=a+b+c=1-f(1)$
  • $\int_0^1h=\alpha$, that is, $\int_0^1g=\frac{a}{3}+\frac{b}{2}+c= \alpha - \beta$

It's easy to check that a solution $(a,b,c)$ to the system above is given by

$$\Big(3-6(\alpha-\beta)-3f(0)-3f(1),-2+6(\alpha-\beta)+4f(0)+2f(1),-f(0)\Big)$$

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