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Consider $f:[-1,1]\to\mathbb{R}$, $x\mapsto \begin{cases} 1, & \text{if } x=0 \\ 0 & \text{else } \end{cases}$ I want to know why f is Riemann integrable and I tried something. First of all we had the definition in lecture: $\int_{-1}^{1^*} f(x)\,dx=inf\{\int_{-1}^1 \phi(x)\,dx;\phi:[-1,1]\to\mathbb{R} \;\text{step function}, f\le\phi \}$ and $\int_{-1^*}^{1} f(x)\,dx=inf\{\int_{-1}^1 \phi(x)\,dx;\phi:[-1,1]\to\mathbb{R} \;\text{step function}, \phi\le f \}$. Now the bounded function f is Riemann integrable, if $\int_{-1}^{1^*} f(x)\,dx=\int_{-1^*}^{1} f(x)\,dx$.

My first try: consider the step function $\phi:[-1,1]\setminus\{0\}\to\mathbb{R},\; x\mapsto 0$. It is $f=\phi$ everywhere on $[-1,1]\setminus \{0\}$. Therefore it is
$\int_{-1}^{1^*} f(x)\,dx=inf\{\int_{-1}^1 \phi(x)\,dx;\phi:[-1,1]\to\mathbb{R} \;\text{step function}, f\le\phi \}=\int_{-1}^{1} \phi(x)\,dx=0$ and $\int_{-1^*}^{1} f(x)\,dx=\int_{-1}^{1} \phi(x)\,dx=0$. Therefore f is Riemann-integrable and $\int_{-1}^{1} f(x)\,dx=0$. Could you help me, to correct my solution?

I'm not sure if it is ok, because my step function isn't defined in $x=0$. But we say that $\phi:[-1,1]\to\mathbb{R}$ is a step function if you have a partition of the interval $[-1,1]$, $x_0=-1<x_1<..<x_n=1$, such that $\phi_{|(x_{i-1},x_i)}=c_i\in\mathbb{R}$ for $i=1,..,n$.
Maybe I have to define $\phi$ in $x=0$ too but take $x=0$ as one of the $x_i's$.

I will try next to do this with Riemann Sums. Regards

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  • $\begingroup$ instead of doing this with your hands, you can prove that the Riemann Integral does not change it's value if you change the function in one point (in fact you can do this for countably many point's) and then you can see that your $f$ is the constant function zero, with the changing one point at zero. $\endgroup$ – sha Apr 24 '15 at 12:30
  • $\begingroup$ Well a function $f: [a,b] \to \mathbb{R}$ is Riemann integrable iff it is bounded and the set of discontinuity points is a zero set (Riemann-Lebesgue Theorem). $\endgroup$ – Jose Antonio Apr 24 '15 at 12:34
  • $\begingroup$ yes, thanks. but I don't know the theorem from lecture and I want to know if it is correct what I have tried $\endgroup$ – Flap Apr 24 '15 at 12:39
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Fix $1> \epsilon > 0$, and consider the step function $$ \varphi(x) = \begin{cases} 1 &: -\epsilon \leq x \leq \epsilon \\ 0 &: \text{ otherwise} \end{cases} $$ Then $$ \int_{-1}^1 \varphi(x)dx = 2\epsilon $$ Hence $$ \int_{-1}^{1^{\ast}} f(x)dx = 0 $$ Now for any step function $\psi \leq f$, then choose a partition $-1 < x_1 < x_2 < \ldots < x_n = 1$ and constants $c_i$ such that $\psi = c_i$ on $(x_{i-1},x_i)$. Now consider the two cases

  • $0 = x_i$ for some $i$ or
  • $0 \in (x_{i-1},x_i)$ for some $i$

In either case, prove that $$ \int_{-1}^1 \psi(x)dx \leq 0 $$ Since the constant function $0 \leq f$ is a step function, it follows that $$ \int_{-1^{\ast}}^{1} f(x)dx = 0 $$ and so $f$ is Riemann integrable.

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A piecewise constant function lies below $f$ if and only if it is nonpositive everywhere. So the supremum of the lower sums of $f$ is zero. Among piecewise constant functions lying above $f$, for any $\delta>0$ we have a function which is $1$ on $[-\delta,\delta]$ and zero elsewhere. The integral of this function is $2 \delta$, which can be made arbitrarily small, so the infimum of the upper sums of $f$ is also zero. We have checked the Darboux criterion for integrability.

What might be left to prove is that the Darboux definition is equivalent to your original definition of Riemann integrability.

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  • $\begingroup$ thank you! If I compare, it wasn't enough what I have done. Thanks $\endgroup$ – Flap Apr 24 '15 at 13:04

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