How do I show that if $f$ is integrable on $[c,b]$ for all $c \in [a,b]$, then $f$ is integrable on $[a,b]$?

Question

Let $a < b$. Suppose that the function $f: [a,b] \to \mathbb{R}$ is bounded and Riemann integrable on $[c,b]$ for every $a < c < b$. Prove that $f$ is Riemann integrable on $[a,b]$ and that

$$\int_{a}^{b} f(x) dx = \lim_{c \to a} \int_{c}^{b} f(x) dx$$

I can see that it is true, but I honestly have no idea how to prove that.

My thought would be to show that the upper and lower sums are equal: $$U(f) = L(f) = \int_{a}^{b} f(x) dx$$ In terms of epsilon-delta, I would want to show that $U(f,P) - L(f,P) \leq \epsilon$ for any $\epsilon > 0$.

Answer 1

Once you establish that $f$ is Riemann integrable on $[a,b]$ (for example as shown by Xander Henderson) there are several ways to proceed in showing that

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

(1) Use the fact that if $f$ is Riemann integrable on $[a,b]$, then it is integrable on any subinterval and

$$\left|\int_a^b f(x) \, dx - \int_c^b f(x) \, dx \right| = \left|\int_a^c f(x) \, dx \right| \leqslant \int_a^c|f(x)| \, dx \\\leqslant \sup_{x \in [a,b]}|f(x)|\cdot (c-a) \underset{c \to a+}\longrightarrow 0$$

This argument obviously invokes several basic properties of the Riemann integral without proof.

For more of a "first principles" proof ...

(2) There exists a partition $P: a = x_0< x_1=c< \ldots < x_n =b$ such that for any Riemann sum $S(P,f) = \sum_{j=1}^n f(\xi_j)(x_j - x_{j-1})$, we have

$$\left|\int_a^bf(x) \, dx-S(P,f)\right|= \left|\int_a^bf(x) \, dx-\sum_{j=1}^n f(\xi_j)(x_j - x_{j-1})\right|< \frac{\epsilon}{3},$$

and by refining the partition if necessary we can assume that with $M = \sup_{x \in [a,b]}|f(x)|$,

$$c-a = x_1-a < \frac{\epsilon}{3M}, \\ \left|\int_c^bf(x) \, dx-S(P',f)\right|=\left|\int_c^bf(x) \, dx-\sum_{j=2}^n f(\xi_j)(x_j - x_{j-1})\right|< \frac{\epsilon}{3}$$

Thus, for every $a< c < a+\frac{\epsilon}{3M}$,

$$ \left|\int_a^bf(x) \, dx-\int_c^bf(x) \, dx \right|\\ \leqslant \left|\int_a^bf(x) \, dx-S(P,f)\right|+ |S(P,f) - S(P',f)|+\left|\int_c^bf(x) \, dx-S(P',f)\right|\\ \leqslant \frac{2\epsilon}{3} + |f(\xi_1)|(x_1-a)\leqslant \frac{2\epsilon}{3} + M(x_1-a)< \epsilon,$$

and, therefore,

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

Answer 2

Since the function is bounded say $|f(x)\le M$, then $|\int _a^cf(x)dx|\le \int _a^c|f(x)|dx\le M(c-a)\to 0$

Answer 3

While there are certainly slicker proofs than what I am about to present, it does not seem unreasonable to work directly from the definitions. My feeling is that an elementary proof gives one a better idea of what the definitions and theorems actually mean, and are therefore, in some respects, more enlightening.

Definitions: Let $f : [a,b] \to \mathbb{R}$ be bounded. Let $$P = \{ a = x_0 < x_1 < \dotsb < x_n = b\} $$ be a partition of $[a,b]$, and for each $j = 1, 2, \dots, n$, define $$ m_j = \inf_{x_{j-1} \le x \le x_j} f(x) \qquad\text{and}\qquad M_j = \sup_{x_{j-1}\le x \le x_j} f(x). $$ The lower and upper Riemann sum subordinate to $P$ are $$ L(f,[a,b],P) = \sum_{j=1}^{n} m_j (x_{j} - x_{j-1}) $$ and $$ U(f,[a,b],P) = \sum_{j=1}^{n} M_j (x_{j} - x_{j-1}), $$ respectively. The lower and upper Riemann integrals are $$ L(f,[a,b]) = \sup \{ L(f,[a,b],P) : \text{$P$ is a partition of $[a,b]$} \} $$ and $$ U(f,[a,b]) = \inf \{ U(f,[a,b],P) : \text{$P$ is a partition of $[a,b]$} \}, $$ respectively. If $L(f) = U(f)$, then we say that $f$ is Riemann integrable on $[a,b]$, and define the Riemann integral to be the common value, i.e. $$ \int_{a}^{b} f(x) \,\mathrm{d}x = L(f) = U(f). $$

Note that for any partition $P$, $$ L(f,[a,b],P) \le L(f, [a,b]) \le U(f,[a,b]) \le U(f,[a,b],P). $$ In order to show that a function is Riemann integrable, it is therefore sufficient to show that $$ U(f,[a,b]) - L(f,[a,b]) \le U(f,[a,b],P) - L(f,[a,b], P) $$ can be made arbitrarily small by choosing an appropriate partition of $[a,b]$. It is also worth noting that if we refine a partition, the upper and lower sums "behave nicely". This is encapsulated in the following lemma, which I will state without proof.

Lemma: Let $P$ be a partition of $[a,b]$, and let $P'$ be a refinement of $P$ (that is, $P'$ contains all of the points in $P$, and possibly more). Then $$ L(f,[a,b],P) \le L(f,[a,b],P') \le U(f,[a,b],P') \le U(f,[a,b],P). $$

The basic idea of the proof is that $$ \bigl( \sup_{x_{j-1} \le x \le x_j} x \bigr) \bigl( x_j - x_{j-1} \bigr) \le \bigl( \sup_{x_{j-1} \le x \le x'} x \bigr) \bigl( x_j - x' \bigr) + \bigl( \sup_{x' \le x \le x_j} x \bigr) \bigl( x_j - x' \bigr) $$ for any $x' \in [x_{j-1}, x_j]$. Thought about carefully, this lets us bound $L(f,[a,b],P)$ from above by $L(f,[a,b],P')$. As similar argument will work for the upper Riemann sums.

In order to prove the desired result, we need to get control on the lower and upper Riemann sums, which will allow us to control the lower and upper Riemann integrals. The essential idea is that we can control the sums on $[a,c]$ by choosing $c$ sufficiently close to $a$, and we can control the sums on $[c,b]$ via the Riemann integrability of $f$ on every such interval.

More precisely, choose $\varepsilon > 0$. As $f$ is bounded, there exist $m$ and $M$ such that $m \le f(x) \le M$ for all $x \in [a,b]$. Chose $x_1 > a$ so that $$ (M-m) (x_1 - a) < \frac{\varepsilon}{2}. $$ Since $f$ is Riemann integrable on $[x_1,b]$, there exists a partition $P_1$ of $[x_1,b]$ such that $$ \int_{x_1}^{b} f(x)\,\mathrm{d}x = L(f,[x_1,b]) < L(f,[x_1,b],P_1) + \frac{\varepsilon}{4}. $$ Note that this follows from the definition of $L(f,[x_1,b])$ as a supremum. Similarly, there is a partition $P_2$ of $[x_1,b]$ such that $$ \int_{x_1}^{b} f(x)\,\mathrm{d}x = U(f,[x_1,b]) > U(f,[x_1,b],P_1) - \frac{\varepsilon}{4}. $$ Take $P' = \{ x_1 < x_2 < \dotsb < x_n = b\}$ to be any common refinement of $P_1$ and $P_2$, e.g. $P' = P_1 \cup P_2$, and define $$ P = \{ a = x_0 < x_1 < \dotsb < x_n = b\}. $$ Note that $P$ is a partition of $[a,b]$, and that \begin{align*} &U(f,[a,b]) - L(f,[a,b]) \\ &\qquad\le U(f,[a,b],P) - L(f,[a,b],P) \\ &\qquad= \sum_{j=1}^{n} M_j (x_{j} - x_{j-1}) - \sum_{j=1}^{n} m_j (x_{j} x_{j-1}) \\ &\qquad= (M-m) (x_1 - a) + \sum_{j=2}^{n} M_j (x_{j} - x_{j-1}) - \sum_{j=2}^{n} m_j (x_{j} x_{j-1}) \\ &\qquad< \frac{\varepsilon}{2} + U(f,[x_1,b],P') - L(f,[x_1,b], P') \\ &\qquad\le \frac{\varepsilon}{2} + U(f,[x_1,b],P_2) - L(f,[x_1,b], P_1) && \text{(by the lemma)} \\ &\qquad< \frac{\varepsilon}{2} + \left( U(f,[x_1,b]) + \frac{\varepsilon}{4} \right) - \left( L(f,[x_1,b]) - \frac{\varepsilon}{4} \right) \\ &\qquad= \varepsilon, \end{align*} with the last equality following from the fact that $L(f,[x_1,b]) = U(f,[x_1,b])$. This completes the proof, as we have shown that the difference between the lower and upper Riemann integrals can be made arbitrarily small by choosing an appropriate partition.

Answer 4

A function $f$ bounded it is Riemann integrable if and only if it is continuous almost everywhere, i.e., the set of discontinuities has measure $0$. Since for all $c>a$, $f$ it is Riemann integrable on $[c,b]$ it follows that the set of discontinuities on $[c,b]$ has measure $0$, therefore on $[a,b]$.