$f$ is bounded by $M$ on $[a, b]$ and if the restriction of $f$ to every interval $[c, b]$ where $c$ in $(a, b)$ is Riemann integrable

If $f$ is bounded by $M$ on $[a, b]$ and if the restriction of $f$ to every interval $[c, b]$ where $c$ in $(a, b)$ is Riemann integrable, then $f$ is Riemann integrable and that $\int _c^b f \to \int_a^b f$ as $c \to a+$.

I am stuck with this problem.

Solution 1:

Recall that a function $f$ is Riemann integrable on $[a,b]$ if and only if it is bounded on $[a,b]$ and its set of discontinuities has Lebesgue measure $0$.

Since $f$ is Riemann integrable on $[c,b]$ for all $c \in (a,b)$, we know that the measure of the discontinuities of $f$ on $[c,b]$ is zero for all $c \in (c,b)$. Since $[a+1/n,b]$ converges upwards to $(a,b]$ we know that the measure of $f$'s discontinuities on $(a,b]$ is $0$, the limit of the measure of $f$'s discontinuous on $[a+1/n,b]$. Finally, possibly adding a single point of measure zero to get account for a discontinuity at $a$ we still have that the measure of $f$'s discontinuities on $[a,b]$ is zero.

Thus $f$ is Riemann integrable in $[a,b]$. The fact that the integrals converge then follows by Lebesgue dominated convergence since the family $1_{[c,b]}(x)f(x)$ is dominated by the integrable constant function $M$.

Solution 2:

The answer by nullUser is good, but uses some sophisticated ideas like an understanding of the relationship between Lebesgue integrability and Riemann integrability. I thought it might be nice to give a direct, straightforward answer as well.

We will show that $f$ is integrable on $[a,b]$ by constructing partitions for the definition of Riemann integration.

So let $\epsilon > 0$ be an allowed error. Denote by $M$ the maximum value of $|f|$ on the interval $[a,b]$.

(In your post, you imply that $f$ needs only an upper bound, but this is insufficient. Consider $-1/x$ on the interval $[0,1]$, defined to be $0$ at $0$, for an instance of a function integrable on any $[\delta, 1]$ for $\delta > 0$, but not integrable on $[0,1]$.)

Now choose $\delta > 0$ such that $M\delta < \frac{\epsilon}{4}$. Fix a partition $P'$ of $[a + \delta, b]$ such that the difference between the upper Riemann bound and lower Riemann bound of $f$ on $P'$ is at most $\frac{\epsilon}{2}$, and which we know exists as $f$ is Riemann integrable.

We construct a partition $P$ for the interval $[a,b]$ from the partition $P'$ for $[a + \delta, b]$ by adding in the single point $a$. We see that the difference between the upper bound Riemann bound and lower Riemann bound for $P$ is at most the greatest difference on $[a, a + \delta]$ added to the greatest difference on $[a + \delta, b]$, which is $$ 2 \frac{\epsilon}{4} + \frac{\epsilon}{2} = \epsilon.$$

So for any $\epsilon$< we have explicitly constructed a partition satisfying the requirements for Riemann integrability.

For the second part: now that we know that $f$ is integrable on $[a,b]$, we know that $$ \int_a^b f = \int_a^c f + \int_c^b f.$$ Since $$ \left \lvert \int_a^c f(x)dx \right \rvert \leq M \lvert c-a\rvert \to 0$$ as $c \to a$, we have that $\int_c^b f \to \int_a^b f$. $\spadesuit$

Solution 3:

Proof. Since $f$ is bounded on $[a,b]$, there exists an $M>0$ such that $|f(x)|≤M$ for all $x∈[a,b]$. Choose an $\varepsilon>0$. Define $δ_1=\min\{\varepsilon/2M,b-a\}$. Choose $c∈(a,b)$ such that $a<c<a+δ_1$. Consider a tagged partition $\dot{\mathcal{P}}_1=\{([x_{i-1},x_i],t_i)\}_{i=1}^n$ of $[a,b]$ with $\Vert \dot{\mathcal{P}}_1\Vert<δ_1$. Then \begin{equation} \left|S(f|_{[a,c]}; \dot{\mathcal{P}}_1)\right|≤\sum_{i=1}^n|f(t_i)|(x_i-x_{i-1})≤M\sum_{i=1}^n(x_i-x_{i-1})=M(c-a)<\frac{\varepsilon}{2} \tag{1}. \end{equation}

This proves that the restriction of $f$ to $[a,c]$ is Riemann integrable with

\begin{equation} \lim_{c\rightarrow a^+} \int_a^c f=0. \end{equation}

Since $f∈\mathcal{R}[c,b]$, there exists a $δ_2>0$ such that for any tagged partition $\dot{\mathcal{P}}_2$ of $[c,b]$ with $\Vert \dot{\mathcal{P}}_2\Vert<δ_2$,

\begin{equation} \left|S(f|_{[c,b]};\dot{\mathcal{P}}_2)-∫_c^b f\right|<\frac{\varepsilon}{2}. \end{equation}

Define $\delta=\min\{\delta_1,\delta_2\}$. Consider a tagged partition $\dot{\mathcal{P}}$ of $[a,b]$ as $\mathcal{P}=\mathcal{P}_1 \cup \mathcal{P}_2$ with the same tags as in $\dot{\mathcal{P}}_1$ and $\dot{\mathcal{P}}_2$. Then $\Vert \dot{\mathcal{P}}\Vert<δ$ and \begin{equation} \left|S(f;\dot{\mathcal{P}})-S(f|_{[c,b]} ;\dot{\mathcal{P}}_2)\right|=\left|S(f|_{[a,c]};\dot{\mathcal{P}}_1)\right|<\frac{\varepsilon}{2}. \end{equation}

Now it follows from the triangle inequality that \begin{equation} \left|S(f;\dot{\mathcal{P}})-∫_c^b f\right|≤\left|S(f;\dot{\mathcal{P}})-S(f|_{[c,b]};\dot{\mathcal{P}}_2)\right|+\left|S(f|_{[c,b]};\dot{\mathcal{P}}_2)-∫_c^b f\right|<\varepsilon. \end{equation} This proves that $f∈\mathcal{R}[a,b]$. By Additivity Theorem, \begin{equation} ∫_a^b f=∫_c^b f+∫_a^c f \end{equation} and eq. $(1)$, one obtains \begin{equation} ∫_a^b f=\lim_{c→a^+}⁡∫_c^b f.\blacksquare \end{equation}