Romanian Mathematical Olympiad

The function $f$ has primitives on the interval $[a,b]$ if and only if the function $g=f+1-f(c)$ has primitives on this interval. So, we can assume that $f(c)=1$. Since $f$ is continuous at $c$, there exists $[\alpha ,\beta ]\subset [a,b]$ such that $c\in (\alpha ,\beta )$ and $0<f(x)<2$, for every $x\in [\alpha ,\beta ]$.

Let $F_a$, respectively $F_b$, be a primitive of $f$ on $[a,c)$, respectively $(c,b]$. Because $F_a$ and $F_b$ are increasing on the intervals $[\alpha ,c)$, respectively $(c,\beta ]$, it follows that the limits ${\lim }_{x\to c}{F}_a(x)=p$ and ${\lim }_{x\to c}{F}_b(x)=q$ exist. Also they are finite, because $F_a$ and $F_b$ are bounded on these intervals (for instance, if $x\in (\alpha ,c)$, then $F_a(x)-F_a(\alpha )=f(\xi )(x-\alpha )$ for some $\xi \in (\alpha ,x)$ and $f$ is bounded on $[\alpha ,x]\subset [\alpha ,\beta ]$).

The function $$F:[a,b]\to \mathbb{R},F(x)=\{\begin{array}{ll}{F}_a(x)-p,& \text{if }x\in [a,c)\\ 0,& \text{if }x=c\\ F_b(x)-q,& \text{if }x\in (c,b]\end{array}$$ is a primitive of $f$ on $[a,b]$.

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Alternative solution.

Alternative Solution. (D. Schwarz) It is enough to show the limits ${\lim }_{x\to c}{F}_a(x)=p$ and ${\lim }_{x\to c}{F}_b(x)=q$ exist and are finite. But $|f|$ is bounded by some value $M$ in a neighborhood $(c-\epsilon ,c+\epsilon )$ of $c$, since $f$ (therefore $|f|$ also) is continuous at $c$.

For $x,x^{\prime }\in (c-\epsilon ,c)$ we will thus have $\frac{F_a(x)-F_a(x^{\prime })}{x-x^{\prime }}=f(\xi )$ at some point $\xi \in (x,x^{\prime })$, hence $|F_a(x)-F_a(x^{\prime })|<M\epsilon$, which implies the existence of the finite limit ${\lim }_{x\to c}{F}_a(x)=p$. An identical reasoning holds for $F_b$.