Let $f$ be a function defined on $(a,b)$, such and integrable on any $[c,d]$ with $a<c<d<b, $ but not integrable in the standard Riemann sense on the entire $[a,b]$. Let me define the function $F$ as an improper integral, namely $$F(u)=\int_{a}^{u}f(x)dx=\lim_{c\rightarrow a}\int_{c}^{u}f(x)dx.$$
Assuming that $F(u)$ is well defined for any $u\in (a,b)$, can we say that $F(u)$ converges to $0$ if $u$ converges to $a$ from the right, meaning $$\lim_{u\rightarrow a}F(u)=0?$$
I know this is true if we consider a standard $f$ that is integrable on the entire $[a,b]$, I am trying to figure out if this holds under more general assumptions. Adding the assumption that $f$ is non-decreasing can be useful?
Let $f$ be Riemann integrable on any interval $[c,d]$ where $a < c < d <b$, and suppose the improper integral over $[a,d]$ denoted as $F(d)$ exists, that is,
$$\tag{1}F(d) = \lim_{c \to a+}\int_c^d f(x) \, dx$$
The improper integral of $f$ over $[a,u]$ must also exist for any $u \in (a,d)$ since for Riemann integrals we have
$$\int_c^u f(x) \, dx = \int_c^d f(x) \, dx - \int_u^d f(x) \, dx,$$
and, hence,
$$\tag{2}F(u) := \lim_{c \to a+}\int_c^u f(x) \, dx = \lim_{c \to a+}\int_c^d f(x) \, dx - \int_u^d f(x) \, dx = F(d) - \int_u^d f(x) \, dx$$
By (1) with $u$ replacing $c$, we have
$$F(d) = \lim_{u \to a+} \int_u^d f(x) \, dx$$
Taking the limit of both sides of (2) as $u \to a+$ we get
$$\lim_{u \to a+}F(u) = \lim_{u \to a+}\left[F(d) - \int_u^d f(x) \, dx \right]= F(d) -F(d) = 0$$