Let $$f:(a,b] \rightarrow \mathbb{R} $$ if the riemann Integral$$ F(c):= \int_c^b f(x) dx $$ for all $$ c \in (a,b) $$ exists and the improper integral $$ F(a)=\int_a^b f(x) dx$$ exists too. Does this mean that F is continuous on [a,b]?
i am completely sure, that this is true, but i need to be sure about this.
On
Consider a counterexample: $$f:(0,1]\longrightarrow \mathbb R$$ $$f(x)=\dfrac{1}{x}$$
Then $\int_c^1f(x)dx=\int_c^1\frac{1}{x}dx=\log 1-\log c= -\log c$. Therefore, the integral exists for all $c\in (0,1)$.
However, $$\int _0^1\frac{1}{x}dx=\lim\limits_{c\rightarrow 0^+}\int_{c}^1\frac{1}{x}dx=\lim\limits_{c\rightarrow 0^+}(-\log c)=+\infty$$ Hence, the integral does not converge. This means that $F(c)=\int_c^1f(x)dx$ is not defined in $c=0$, so it cannot be continuous there.
$F$ must be continuous at $a$ pretty much by definition. $\int_a^b f(x)\ dx$ is defined as $\lim\limits_{c\to a} \int_c^bf(x)\ dx$. In other words, $F(a) = \lim\limits_{c \to a} F(c)$, which means precisely that $F$ is continuous at $a$.