improper integral and dominated convergence Suppose $f:(a,b] \to \mathbb R$ is a continuous function and $\lim_{x\to a+} f(x)=+\infty$

Suppose $f(x)\leq g(x)$ for all $x\in(a,b]$, for some $g：[a,b]\to \mathbb R$ satisfying  $\int_a^b g(x)dx&lt;+\infty$

Can we conclude that

$\lim_{t\to a^+}\int_t^b f(x)dx \leq \int_a^b g(x)dx$  ?

Thanks.
Yes, we can. First, for sufficiently small $h&gt;0$, we have
$$\int_{a+h}^bf(x)\,dx\leq\int_{a+h}^bg(x)\,dx$$
and since the r.h.s tends to a limit as $h\to 0$, we conclude that the l.h.s is
bounded. Moreover, since $f(x)\to\infty$ as $x\to a^+$, we can choose $h&gt;0$
so that $f(x)$ is positive in $(a,a+h)$, and in particular, the integral $\int_{a+h}^bf(x)\,dx$ increases as $h$ decreases to zero. Thus, the function
$h\to\int_{a+h}^bf(x)\,dx$ is monotonically increasing and bounded as $h\to 0^{+}$, hence the limit exists, and obviously is not larger than the integral of $g$.
Yes, we can. First, for sufficiently small $h&gt;0$, we have
$$\int_{a+h}^bf(x)\,dx\leq\int_{a+h}^bg(x)\,dx$$
and since the r.h.s tends to a limit as $h\to 0$, we conclude that the l.h.s is
bounded. Moreover, since $f(x)\to\infty$ as $x\to a^+$, we can choose $h&gt;0$
so that $f(x)$ is positive in $(a,a+h)$, and in particular, the integral $\int_{a+h}^bf(x)\,dx$ increases as $h$ decreases to zero. Thus, the function
$h\to\int_{a+h}^bf(x)\,dx$ is monotonically increasing and bounded as $h\to 0^{+}$, hence the limit exists, and obviously is not larger than the integral of $g$.