prove or disprove on bounded variation

Question

Prove or disprove :

If a function $f$ is bounded on $[\alpha,\beta]$ and Riemann integrable on $[\alpha,\beta]$ then it is integral function $F(x)= \int_{\alpha}^{x} f(t) dt$ is of bounded variation on $[\alpha,\beta]$.

My attempt:

I tried to find a counter example on a function $F(x)$ which is bounded on $[\alpha,\beta]$ and continuous but it's not bounded variation on $[\alpha,\beta]$ and it's derivative is bounded on $[\alpha,\beta]$ and Riemann integrable But almost I didn't find any example satisfying the above conditions.

Answer 1

$F(x)$ must be of bounded variation thats why you couldnt find any counterexample. To see this, observe that if $|f(t)|\leq M$ for every $t\in [\alpha,\beta]$ then, for every $x,y$ \begin{align} |F(x)-F(y)|&=\biggl|\int_{\alpha}^x f(t)\,dt-\int_{\alpha}^y f(t)\,dt\biggr|\\ &=\biggl|\int_{x}^y f(t)\,dt\biggr|\\ &\leq \int_{x}^y |f(t)|\,dt\\ &\leq M|x-y| \end{align} This shows that $F$ is Lipschitz continuous with constant $M$. Now using the definition of bounded variation try to prove that every Lipshcitz continuous function on $[\alpha,\beta]$ is of bounded variation.

Answer 2

Since $f$ is bounded on $[a,b],$ there exists $M > 0$ such that $$ \vert f(t) \vert < M \qquad \forall t \in [a,b]$$ Therefore $F$ has bounded variation since \begin{align*} \sum_{i = 0}^{n-1} \vert F(x_{i+1}) -F(x_i)\vert & = \sum_{i=0}^{n-1}\left \vert \int_{x_i}^{x_{i+1}} f(t)dt\right\vert \\ &\leq \sum_{i = 0}^{n-1} \int_{x_i}^{x_{i+1}} \vert f(t)\vert dt \\ &\leq \sum_{i = 0}^{n-1} \int_{x_i}^{x_{i+1}} M dt \\ &= M \int_a^b dt \\ &= M(b-a) \end{align*} for any partition $P = \{x_0,...,x_n \}$ of $[a,b]$.