Seemingly Obvious Improper Integral Property

Question

The following seems extremely obvious, so much so that I cannot see how to prove it:

If $f:[a,b]\to\mathbb{R}$ is Riemann integrable then $$\int_a^bf(x)dx=\lim_{c\to b^-}\int_a^cf(x)dx$$

Answer 1

Let's use the definition. Let $\epsilon > 0$. We want $\delta > 0$ such that if $b-\delta < x < b$, then $$\left|\int_a^bf(t)\,{\rm d}t-\int_a^xf(t)\,{\rm d}t\right| <\epsilon.$$Since $f$ is Riemann-integrable, $f$ is bounded almost everywhere by a positive constant, say, $M>0$. Choose $\delta = \epsilon/M > 0$ and note that: $$\left|\int_a^bf(t)\,{\rm d}t-\int_a^xf(t)\,{\rm d}t\right| = \left|\int_x^bf(t)\,{\rm d}t\right|\leq \int_x^b|f(t)|\,{\rm d}t\leq M(b-x) < M\delta = M\frac{\epsilon}{M}=\epsilon.$$

Answer 2

Let $\epsilon>0$ be given. Since $f$ is Riemann integrable, then it is bounded on $[a,b]$. Let $M=\sup_{x\in[a,b]}|f(x)|=M$. Then, we have

$$\begin{align} \left|\int_{b-\delta}^{b}f(x)\,dx\right|&\le \int_{b-\delta}^{b}|f(x)|\,dx\\\\ &\le M\,\delta\\\\ &<\epsilon \end{align}$$

whenever $\delta<\epsilon/M$.

Answer 3

Let $m$ and $M$ be lower and upper bounds for $f$ on $[a,b]$. Then the inequality $$m(b-c)\leq\int_c^b f(x) dx \leq M(b-c)$$ is valid for all $c \in [a,b]$, which implies that $$\lim_{c \to b^-} \int_c^b f(x) dx = 0$$ Consequently, $$\begin{aligned} \lim_{c\to b^-} \int_a^c f(x) dx &= \int_a^b f(x) dx - \lim_{c \to b^-} \int_c^b f(x) dx \\ &= \int_a^b f(x) dx \end{aligned}$$