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Here's a problem from my text that's giving me some troubles...I can get pretty close in the proof but something still seems to be missing. Any ideas on where to go with this?

Let $f:[−1,1]\to \Bbb{R}$ denote the function $f(x)=1$ if $0 ≤ x≤ 1$, and $f(x)=0$ otherwise. Prove that $f$ is Riemann integrable that $\int f(x)dx = 1$.

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    You probably meant $f:\mathbb{R}\to\mathbb{R}$. – 2017-02-03
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    hint: if you mean the improper Riemann integral, take any partition $P$ of the interval $[-M,M]$ with $M\ge 1$. By passing to a refinement you can assume $x=1$ and $x=-1$ are included in $P$. – 2017-02-03

2 Answers 2

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You can always use the Darboux definition to help you prove that it is integrable. Remember the darboux definition says that,

if

$$\sup(L(f,p), \forall \text{ partitions of [-1,1]}) = \inf(U(f,p), \forall \text{ partitions of [-1,1]}), \text{ then, $f$ is integrable on that partition}$$

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Choose a Partition $P=\big\{x_1,x_2 ...,x_{n+1}\big\}$,$P_i=[x_i,x_{i+1}]$ of the set [-1,1] with refinement $\Delta P$. Then we have: $$ \sum_{i=1}^n (\sup_{x \in P_i}f(x)-\inf_{x \in P_i}f(x))(x_{i+1}-x_i) \leq 1 \cdot\Delta P $$ since there is only 1 $[x_i,x_{i+1}]$ such that $x_i \leq 0$ and $x_{i+1} \geq 0$ and therefore only 1 term where $\sup_{x \in P_i}f-\inf_{x \in P_i}f \neq0$. Letting $\Delta P \to 0$ gives you the result that f is indeed integrable.
Now to calculate the value, you have that: $$ \int_{-1}^1f(x)dx=\sum_{i=1}^n\sup_{x \in P_i}f(x)(x_{i+1}-x_i) \leq 1+\Delta P \\ \int_{-1}^1f(x)dx=\sum_{i=1}^n\inf_{x \in P_i}f(x)(x_{i+1}-x_i) \geq 1-\Delta P $$ Since, again, the intervall where $\sup$ and $\inf$ do not coincide is at most of length $\Delta P$. Applying the sandwich lemma and $\Delta P \to 0 $ gives you the result.