Let $f:[0,1]\rightarrow \mathbb{R}$ be a bounded function whose restriction to $(0,1]$ is continuous. Prove that $f$ is Riemann integrable.
There are plenty of definitions, worked examples and explanations on inclusive boundaries i.e. $[a,b]$.
But because of the exclusive boundary I don't think I could use $0$ as an infimum. Does the slight difference in the boundary make $f$ a non-uniformly continuous function? If so, how do I prove this differently so the argument still holds?
Any help will be greatly appreciated!
Since $f$ is bounded there exist $m,M$ such that $m \leqslant f(x) \leqslant M$ for all $x\in [0,1]$. For any $\epsilon > 0$, take a point $x_1 \in (0,1)$ such that $(M-m)x_1 < \epsilon/2$.
For partition $P: 0< x_1< x_2< \ldots < x_n = 1$, let $M_j =\sup_{x \in [x_{j-1},x_j]}f(x)$ and $m_j =\inf_{x \in [x_{j-1},x_j]}f(x)$.
We have $(M_1-m_1)x_1 \leqslant (M-m)x_1 < \epsilon/2$, and
$$U(f,P) - L(f,P) = (M_1-m_1)(x_1-0) + \sum_{j=2}^n(M_j-m_j)(x_j - x_{j-1})\\ < \epsilon/2 + \sum_{j=2}^n(M_j-m_j)(x_j - x_{j-1})$$
I'll let you finish by showing that the partition points $x_2,\ldots, x_{n-1}$ can be chosen such that
$$\sum_{j=2}^n(M_j-m_j)(x_j - x_{j-1}) < \epsilon/2$$