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    Hi,
    Define g:[a,b] -> \mathbb{R} by g(x) = a for x=t and g(x) = 0 elsewhere.

    The question is to show that \int^s_r g = 0.

    As I'm learning Riemann integration and the associated content, the following way is the way I need to do it.

    Well, first I showed that g is Riemann integrable by using the Riemann criterion and I considered the partition \{r, t-d, t+d, s\} for small d.

    Now I tried to work out the upper and lower sums. Now I figured that all partitions of [r,s] could be written as \{r, t-d, t+d, s\} (for probably obvious reasons), so the Riemann lower sum (which is the supremum of lower sums of all partitions) is the supremum of the set \{2da : 'small' d\}.

    Now obviously this Riemann lower sum isn't zero. So what's gone wrong?

    (Can someone tell me how I make my first bit of latex look like what's it meant to be? Thanks).
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    (Original post by Usar)
    Now I tried to work out the upper and lower sums. Now I figured that all partitions of [r,s] could be written as \{r, t-d, t+d, s\} (for probably obvious reasons)
    Not really - it's not true!

    (Original post by Usar)
    so the Riemann lower sum (which is the supremum of lower sums of all partitions) is the supremum of the set \{2da : 'small' d\}.
    Is a positive or negative (or zero) here? You'll need to split this up into cases.

    For example, the upper sum of this partition, if a > 0, is obviously just 2da; the lower sum is 0 (because there are points in (t-d, t+d) where g vanishes).

    (Original post by Usar)
    (Can someone tell me how I make my first bit of latex look like what's it meant to be? Thanks).
    \to = \to
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    (Original post by generalebriety)
    Not really - it's not true!
    Are you sure? If x and y are points in the partition and t \not\in [x,y] then the contribution of the rectangle between x and y is 0, so you could just boil it all down to the partition I gave.

    Thanks for the rest of the post, I'll work it at tomorrow when I have more energy.
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    (Original post by Usar)
    you could just boil it all down to the partition I gave.
    That's not what you said, though.

    Anyway, what if I want to consider a partition like {r, t-d, t+2d, s}? Point is, what you said was wrong, even if your intention was right.
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    Thanks for the help -- I solved it.

    I guess a partition like {r, t-q, t+p, s} solve the problem and cover all areas. Is that how you would approach it?
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    What, precisely, is your definition of the Riemann integral?
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    Well, the lower sum L and the upper sum U of a function f:[a,b] -> R with partition P are defined as:
    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    L(P,f) = \sum^{n-1}_{j=0}{m_j(x_{j+1} -x_j)

    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    U(P,f) = \sum^{n-1}_{j=0}{M_j(x_{j+1} -x_j)

    where the xjs are points in a partition.

    And:
    m_j = inf\{f(x) : x_j \leq x \leq x_{j+1}\}
    M_j = sup\{f(x) : x_j \leq x \leq x_{j+1}\}

    Then the lower and upper Riemann sums are:
    L^b_a(f) = sup\{L(P,f) : all partitions P\}
    U^b_a(f) = inf\{L(P,f) : all partitions P\}

    And when f is Riemann integrable, the Riemann sums are equal and are equal to the Riemann integral.
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    So, you need to do two things.

    Show the lower sum is 0 (which should be straightforward).

    Show the upper sum is 0. To do this, show that given \epsilon &gt; 0 you can find a partition with upper sum < epsilon. Then U_a^b(f) &lt; \epsilon \forall \epsilon &gt; 0 and so it must be zero.
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    (Original post by DFranklin)
    So, you need to do two things.

    Show the lower sum is 0 (which should be straightforward).

    Show the upper sum is 0. To do this, show that given \epsilon &gt; 0 you can find a partition with upper sum < epsilon. Then U_a^b(f) &lt; \epsilon \forall \epsilon &gt; 0 and so it must be zero.
    Though just to clarify, as I said earlier, this is for a >= 0, and you'll have to do a < 0 separately (or at least allude to the fact that you've noticed you need to do so).
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    Thanks for the help.
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    Ok, got another problem:
    if f:[a,b] \to \mathbb{R} is continuous, show that for each p > 0, there exists a natural number n so that, for the partition P_n = \{a + k(b-a)/n : k = 0,...,n\}, we have
    |U(P_n, f) - L(P_n, f)| &lt; p.

    (Where U and L are defined as in an above post)

    Now, I've got it down to:
    |U-L| = \frac{(b-a)}{n}|M_0 - m_0 + M_1 - m_1 + ... + M_{n-1} - m_{n-1}|.

    I can't seem to find a suitable n -- not sure how to deal with the sum of M and ms. I tried a few things (triangle rule, max of f(x)) but they were wrong or fruitless. Any help?
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    Does it help you to know that, if f is continuous on a closed, bounded interval, then f is bounded and attains its bounds on that interval?
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    Probably -- I was under the impression that f(x) may shoot off to |infinity|. But of course if f(x) on [a,b] is continuous then it has a value at every x in [a,b]. Thanks for the hint.
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    GE: Surely a bigger help is that if f is cts on a closed bounded interval, it's uniformly cts on that interval too.
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    (Original post by DFranklin)
    GE: Surely a bigger help is that if f is cts on a closed bounded interval, it's uniformly cts on that interval too.
    Yep, but (at least if the current Cambridge course is anything to go by) uniform continuity is normally taught after Riemann integration. I think you can deduce all the relevant properties from my hint...?
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    ^ Not sure I can actually (can't really incorporate max(f) anywhere that I can see). I'll take a look at the uniform continuity property.
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    (Original post by generalebriety)
    Yep, but (at least if the current Cambridge course is anything to go by) uniform continuity is normally taught after Riemann integration. I think you can deduce all the relevant properties from my hint...?
    I don't think boundedness is enough: somewhere or other you need the concept that "f can't vary much over a small interval", and if you are forcing the intervals to be of uniform size then that seems pretty identical to uniform continuity.

    Tired and drunk, so might be more lucid tomorrow...
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    Thanks -- it worked.

    Got another question for anyone (sorry):

    I'm trying to show that the sequence given by x_n = \frac{\pi}{t + 2n\pi} doesn't converge to 1 (I've shown it converges to 0). But I thought I'd plug in the def. using the limit as 1 just to see. (By the way, t is fixed).

    \left|\frac{\pi}{t + 2n\pi} - 1\right| = \left|\frac{\pi - t - 2n\pi}{t + 2n\pi}\right| &lt; \epsilon

\implies

|\pi - (t + 2n\pi)| &lt; \epsilon|t + 2n\pi|

    Now if I choose n s.t t+2n\pi &gt; 0, then the modulus sign on the RHS can be got rid, and if I choose n s.t t+2n\pi &gt; \pi also, then the modulus on the LHS can be got rid of after multiplying the expression by -1.

    After rearranging, this gives n &gt; \frac{t(1-\epsilon) - \pi}{2\pi(\epsilon - 1)}. So if choose n to be the max of these two inequalities, it seems to work. So what have I done wrong?
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    Post your full working.
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    (Original post by DFranklin)
    Post your full working.
    Ah yeah, realised the RHS denominator is negative so the > should be <. Thanks.
 
 
 
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