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For a function to be Rienmann integral, its upper integral and lower integral need to be the same. I don't understand the function they have given here where $x=r/2^n$ thus don't know how to apply to values of the bounds to the functions. Also how to work out the upper and lower sum from this?

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    Twice you wrote it incorrectly: it is **Riemann** . – 2017-01-13
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    @K_uddin I've edited to focus on $f$ as you've mentioned in your title, and retyped the question and kept the original image linked. Just in case, you were asking about $f$ and not $f_n$ I presume? Although I imagine both questions will have a some similarities in their analysis. – 2017-01-13

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Hint: for each fixed $n$ there is only a finite number of points of $x\in[0,1]$ s.t. $x = r/2^n$, $r\in\Bbb Z$. But the set of points $r/2^n\in[0,1]$ for some $r,n\in\Bbb Z$ is dense in $[0,1]$ (and in any subinterval).

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For any fixed $n$ the function is zero on the whole interval up to $0,1/2^n,2/2^n,\dots, (2^n-1)/2^n,1$, where its value is $1$.

For example is $n=3$, then your function $f_3$ is $1$ at $0,1/8,2/8,3/8,4/8,5/8,6/8,7/8,8/8$ and zero everywhere else.

$f$ looks basically like $f_n$ but has more peaks: it is $1$ where $x=r/2^n$, $r,n\in\mathbb{N}$. So it is $1$ at $0,3/4,7/8,10/1224\ $ but not at $5/7,9/10,3/6\ $ or any irrational number.

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    Note that we are not talking about multiple functions parametrized by $n$ here. – 2017-01-13
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    @Vik78 Well there was an $f_n$ in the original attached image as you can still check, but since the o.p. mentioned $f$ and not $f_n$ I presumed they meant $f$. I retyped the question to make it clearer which one the question was presumed to be about. – 2017-01-13
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    Exactly @snulty. I've answered how to interpret the first function $f_n$ of the first question. – 2017-01-13
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    I can roll the edit back until the op responds, since both of the present answers address part (a). – 2017-01-13