We know that if $f:\mathbb{R}\to\mathbb{R}$ is a function such that $$\sup_{x\in\mathbb{R}}|f(y)|<\infty,$$ then the function $g(x)=\int_0^xf(y)dy$ is uniformly continuous.
I am just wondering what can we say about $g$ if $f$ satisfies the following weaker condition $$\sup_{x\in\mathbb{R}}\int_x^{x+1}|f(y)|dy<\infty.$$ Is $g$ uniformly continuous ?
No, $g$ need not be uniformly continuous.
For $n>1$, let $f_n$ be the non-negative continuous function with value $0$ off the set $[n-1/n, n+1/n]$ whose graph contains the straight line segments connecting the points $(n-1/n, 0)$, $(n,n)$, and $(n+1/n,0)$.
Let $f=\sum\limits_{n=2}^\infty f_n$. Then $f$ satisfies your integral condition. But for each $n>2$, $g(n+1/n)-g(n-1/n)=1$; so $g$ is not uniformly continuous.