Prove that if $f$ is bounded in $[a,b]$ and integrable in each interval $[c,b]$ with $a<c<b,$ then $\int_a^b f =\lim_{c\to a+}\int_c^b f$

Question

Prove that if $f$ is bounded in $[a,b]$ and integrable in each interval $[c,b]$ with $a<c<b,$ then $f$ is integrable in $[a,b]$ and also
$$\int_a^b f =\lim_{c\to a+}\int_c^b f$$

I already proved the integrability of $f$:

Given $\varepsilon>0$ we should find a partition $P$ such that $U(f,P)-L(f,P)<\varepsilon$

$$U(f,P)-L(f,P)=\sum_{k=1}^n(M_k-m_k)\Delta x_k=(M_1-m_1)(x_1-a)+\sum_{k=2}^n(M_k-m_k)\Delta x_k$$
I have to find $x_1$ small enough such that $(M_1-m_1)(x_1-a)<\frac{\varepsilon}{2}$.

$f$ is bounded, $\implies \exists M>0$ such that $|f(x)|\leq M \ \forall x \in [a,b].$ And $M_1-m_1\leq2M$. With the hypothesis, $f$ is integrable in $[x_1,b] \implies \exists P_1$ partition of $[x_1,b]$ with $U(f,P_1)-L(f,P_1)<\frac\varepsilon2$.

If we have a partition $P=\{a\}\cup P$ of $[a,b]$ such that $U(f,P)-L(f,P)\leq 2M(x_1-a)+U(f,P1)-L(f,P_1)<\frac\varepsilon2+\frac\varepsilon2=\varepsilon$, so, $f$ is integrable in $[a,b]$

How can I find that $$\int_a^b f =\lim_{c\to a+}\int_c^b f$$

Answer 1

Because $f$ is integrable in $[a,b]$ we have $$\int_a^b f \, dx = \int_a^c f \, dx + \int_c^b f \, dx$$ but $$0 \le \lim_{c \to a+} \left|\int_a^c f \, dx\right| \le \lim_{c \to a+}\int_a^c |f| \, dx \le \lim_{c \to a+}M(c-a) = 0$$ hence $$\lim_{c \to a+}\int_a^c f \, dx = 0$$ and so in total we get:

$$\int_a^b f \, dx = \lim_{c \to a+} \int_a^b f \, dx = \lim_{c \to a+}\int_a^c f \, dx + \lim_{c \to a+}\int_c^b f \, dx = \lim_{c \to a+}\int_c^b f \, dx$$