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I got this question which popped up in my mind:

Prove or disprove:

If a function $f$ is bounded in a closed interval $[a,b]$ and if for each $x\in(a,b)$, $f$ is integrable over $[x,b]$, Then $f$ is integrable over $[a,b]$

Note: This question rose from the following problem that I've come across:

Let $f$ be a function defined on the interval $[0,1]$ that satisfy $\forall x\in[0,1], 4<f(x)<5$. If for each $x\in(0,1)$, $f$ is integrable over $[x,1]$, Then $f$ is integrable over $[0,1]$.

Here's my answer:

We must show that for each $0<\epsilon$ there exist a partition $P$ of the interval $[0,1]$ such that $S(P)-s(P)<\epsilon$ (Where $S(P)$ is the upper sum and $s(P)$ is the lower sum).

Lets take $\epsilon\in(0,2)$, Then $\frac{\epsilon}{2} \in (0,1)$ which implies that $f$ is integrable over $[\frac{\epsilon}{2},1]$, and hence we get that for each $0<\kappa$ there exist a partition $Q$ of the interval $[\frac{\epsilon}{2},1]$ such that $S(Q)-s(Q)<\kappa$.

Now because $0<\frac{\epsilon}{2}$, we get that there exist a partition $Q$ of the interval $[\frac{\epsilon}{2},1]$ such that $S(Q)-s(Q)<\frac{\epsilon}{2}$

If we take $P=Q\cup \{0\}$, we get that $P$ is a partition of $[0,1]$ And hence we get that $ S(P)-s(P)=(sup f([0,\frac{\epsilon}{2}] - inf f([0,\frac{\epsilon}{2}])*\frac{\epsilon}{2}+S(Q)-s(Q)$

Now because $\forall x\in[0,\frac{\epsilon}{2}], 4<f(x )<5$, we get that $sup f([0,\frac{\epsilon}{2}]\leq 5$ and $4\geq inf f([0,\frac{\epsilon}{2}])$, And so $sup f([0,\frac{\epsilon}{2}] - inf f([0,\frac{\epsilon}{2}])\leq 1$ Which implies that $(sup f([0,\frac{\epsilon}{2}] - inf f([0,\frac{\epsilon}{2}]))*\frac{\epsilon}{2}\leq \frac{\epsilon}{2}$, and because $S(Q)-s(Q)<\frac{\epsilon}{2}$, We get that $(sup f([0,\frac{\epsilon}{2}] - inf f([0,\frac{\epsilon}{2}])*\frac{\epsilon}{2}+S(Q)-s(Q)<\epsilon$ And so $S(P)-s(P)<\epsilon$.

Therefore. We've shown that for each $\epsilon\in(0,2)$ there exist a partition $P$ of $[0,1]$ such that $S (P)-s(P)<\epsilon$.

Now if $\epsilon\in[2,\infty)$ we can take the partition that corresponds to $\frac{1}{2}$ and get that there exist a partition $P$ of $[0,1]$ such that $S (P)-s(P)<\frac{1}{2}<2\leq\epsilon$.

And so we've shown that for each $0<\epsilon$ there exist a partition $P$ of $[0,1]$ such that $S(P)-s(P)<\epsilon$ Which implies that $f$ is integrable over $[0,1]$ as was to be shown.

Now how can we generalize this result to the original question ?

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  • $\begingroup$ There is something called "real induction". It's like induction, but real. You can google it. $\endgroup$
    – Pedro
    Sep 6, 2014 at 19:38

2 Answers 2

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Your proof is correct. To generalize it, give names $M$ and $m$ to the sup and inf of $f$ over $[a,b]$. Then break the interval into $I_1 = [a,a+\alpha]$ and $I_2 = [a+\alpha,b]$, where $\alpha = \min(b-a,\epsilon/2(M-m))$. Choose a partition Q of $I_2$ for which $S(Q) - s(Q) < \epsilon/2$, then continue as above.

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Your premise is that "for each $x \in (a,b), f$ is integrable over $[x,b]$". So you can take an $x$ arbitrarily close to $a$, which shows that $f$ is integrable over $\left(a,b\right]$. This differs from the domain of $f$ by the singleton $\{a\}$. Therefore, $f$ is also integrable over $[a, b]$.

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