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Are these conditions sufficient to demonstrate integrability of the function? Or did I miss writing down integrability as hypothesis

Robert Shore
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Marco
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  • You have to assume integrability. – Kavi Rama Murthy Aug 31 '21 at 23:49
  • If you assume $f$ is continuous on $[a,b]$, not just at a single point in the interval, then it is automatically integrable. Also note that you shouldn't quantify over the constant $a$, you can only quantify over variables. – David Sheard Aug 31 '21 at 23:53
  • There is no point in stating $a \in [a,b]$. – copper.hat Sep 01 '21 at 00:00
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    As others have mentioned, $f$ isn't necessarily integrable. For example, consider the function $$g(x)=\begin{cases} x,&\text{if }x\in\Bbb Q\ 0,&\text{if }x\in\Bbb R\setminus\Bbb Q; \end{cases}$$ which is continuous only at $x=0$. The restriction of $g(x):=f(x-c)+1$ to $[a,b]$ satisfies your conditions but isn't integrable on $[a,b]$. – Mandelbrot Sep 01 '21 at 00:18
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    @Mandelbrot Your function is Lebesgue integrable because it's constant outside a set of measure $0$. – Robert Shore Sep 01 '21 at 00:55
  • Yes, you are correct. However, I assumed that OP was talking about the Riemann Integral since his question is a standard homework problem in this context. I guess one can always redefine my function to include the characteristic function of some vitali set on some portion of $[a,b]$ in order for it to be not integrable.

    There is also another mistake in my answer. It should be

    $$f(x):=g(x-c) +1.$$

     – Mandelbrot Sep 01 '21 at 01:01
  • You can just change Q to some non-measurable set and it won't be Lebesgue integrable. – A. Thomas Yerger Sep 01 '21 at 01:06

1 Answers1

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Provided that $f$ is integrable on $[a,b]$, you can deduce the inequality $\int_{a}^{b}f(x)dx>0$ as follows:

We will have either $c=a$, $c\in(a,b)$, or $c=b$. I will only argue the case $c\in(a,b)$, since the proof for the other cases is simply a matter of replacing the normal $(\varepsilon,\delta)$ definition of a limit with its one-sided variant.

Since $f$ is continuous at $c$, we know that for any $\varepsilon>0$, there is a $\delta>0$ such that for every $x\in[c-\delta,x+\delta]$,

$$\left|f(x)-f(c)\right|<\varepsilon$$

In particular, this is true of $f(c)$, since $f(c)>0$. Thus, there is a real number $\delta_1>0$ such that for every $x\in[c-\delta_1,c+\delta_1]$,

$$\left|f(x)-f(c)\right|<f(c)$$

This inequality is equivalent to $0<f(x)<2f(c)$, showing that $f$ is strictly positive over the interval $[c-\delta_1,c+\delta_1]\subseteq [a,b]$. We deduce that the integral

$$\int_{c-\delta_1}^{c+\delta_1}f(x)dx$$

will exist and also be strictly positive. To leverage this result, split the integral $\int_{a}^{b}f(x)dx$ into a sum of integrals as follows:

$$\int_{a}^{b}f(x)dx = \int_{a}^{c-\delta_1}f(x)dx+\int_{c-\delta_1}^{c+\delta_1}f(x)dx+\int_{c+\delta_1}^{b}f(x)dx$$

Since $f(x)\geq 0$ over $[a,c-\delta_1]$ and $[c+\delta_1,b]$, the integrals $\int_{a}^{c-\delta_1}f(x)dx$ and $\int_{c+\delta_1}^{b}f(x)dx$ are both at least zero. It follows that their sum is at least zero, so

$$\int_{a}^{c-\delta_1}f(x)dx+\int_{c+\delta_1}^{b}f(x)dx\geq 0$$

Adding $\int_{c-\delta_1}^{c+\delta_1}f(x)dx$ to both sides of this inequality, we get

$$\int_{a}^{c-\delta_1}f(x)dx+\int_{c-\delta_1}^{c+\delta_1}f(x)dx+\int_{c+\delta_1}^{b}f(x)dx\geq \int_{c-\delta_1}^{c+\delta_1}f(x)dx>0$$

so

$$\int_{a}^{b}f(x)dx\geq\int_{c-\delta_1}^{c+\delta_1}f(x)dx>0$$

This implies that $\int_{a}^{b}f(x)dx>0$, so we're done.

Alann Rosas
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