Does Riemann integral of everywhere continuous and nowhere differentiable functions (with chosen values at the boundary points) can attain any value?

Question

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.

Answer 1

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.

Answer 2

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

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