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Math Help - A few analysis questions

  1. #1
    MHF Contributor Mathstud28's Avatar
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    A few analysis questions

    Hey everyone! Any input would be appreciated.

    A note on notation, f\in\mathfrak{R}(\alpha) on [a,b] means that f is Riemann-Stieltjes integrable with respect to \alpha(x) on the interval [a,b]

    Question 1: Let x_0\in[a,b] and define f on [a,b] by f:x\longmapsto\left\{ \begin{array}{rcl} 0 & \mbox{if} & x\ne x_0\\ c & \mbox{if} & x=x_0 \end{array} \right.. Also, let \alpha(x) be continuous at x=x_0. Is f\in\mathfrak{R}(\alpha) on [a,b]? If so what is its value.

    Answer: Using the following lemma:

    Lemma: Suppose that f is bounded on [a,b], f has a finite number of discontinuities and \alpha is continuous at all those points, then f\in\mathfrak{R}(\alpha) on [a,b].

    We can see that f\in\mathfrak{R}(\alpha) since f has only one discontinuity, at x_0, and \alpha is continuous there.

    Next to find the value of this integral consider L\left(P,f,\alpha\right) for any partitions. Now consider the interval [x_{i-1},x_i] created by P. Since P is a finite point set, it follows that [x_{i-1},x_i] has infinite points. Which indicates it contains values other than x_0. So for any interval on P we can see that m_i=\inf_{[x_{i-1},x_i]}f=0. Now since L\left(P,f,\alpha\right)=\sum_{i=1}^{n}m_i\Delta \alpha_i we can see that for any partition L\left(P,f,\alpha\right)=0. Now since f\in\mathfrak{R}(\alpha) it follows that \int_a^b f~d\alpha=\inf_{P}L\left(P,f,\alpha\right)=0
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    Question 2: Let f:x\longmapsto \left\{ \begin{array}{rcl} 0 & \mbox{if} & x\not\in\mathbb{Q}\\ 1 & \mbox{if} & x\in\mathbb{Q} \end{array} \right., this is known as Dirchlet's function. Show that f\not\in\mathfrak{R}(\alpha) for any a<b

    Answer: This one is simple I think. Consider any partition P of [a,b]. Now consider any interval [x_{i-1},x_i] formed by P. For the same reason as above [x_{i-1},x_i] is infinite. Now since R-\left\{\mathbb{Q}\right\} is dense in any subset of the reals it follows that there are value of x in [x_{i-1},x_i] such that x\not\in\mathbb{Q} where it follows similarly to above that L\left(P,f,\alpha\right)=0 for any partition. Now following a similar argument noting that \mathbb{Q} is dense in any subset of the reals we can see that for any partition U\left(P,f,\alpha\right)=1. So for any partition U\left(P,f,\alpha\right)-L\left(P,f,\alpha\right)=1. Where it follows that f\not\in\mathfrak{R}(\alpha) since it fails the criterion that given any \varepsilon>0 we must have a partition such that U\left(P,f,\alpha\right)-L\left(P,f,\alpha\right)<\varepsilon

    --------------------------------------------------------------------
    Question 3: Does f^2\in\mathfrak{R}(\alpha)\implies f\in\mathfrak{R}(\alpha)?

    Answer: No. Counterexample f:x\longmapsto\left\{ \begin{array}{rcl} -1 & \mbox{if} & x\not\in\mathbb{Q}\\ 1 & \mbox{if} & x\in\mathbb{Q}\end{array}\right.. The reason is that f^2:x\longmapsto\left\{ \begin{array}{rcl} 1 & \mbox{if} & x\not\in\mathbb{Q}\\ 1 & \mbox{if} & x\in\mathbb{Q}\end{array} \right.=1 which is continuous thus integrable, but for the same reason as f in question 2 f is not integrable.

    ------------------------------------------------------------------------
    Question 4: Suppose that f\geqslant 0, f is continuous on [a,b] and \int_a^b f~dx=0. Prove that f(x)=0~~\forall x\in[a,b].

    Answer: It is obvious that if f=0\implies \int_a^b f~dx=0. So instead let us prove that if f>0\implies \int_a^b f~dx>0.

    Since f\ne 0 we can find a Partition of [a,b] such that for some interval [x_{i-1},x_i] we have that f>0~~\forall x \in[x_{i-1},x_i], now since this interval is closed it follows that \inf_{[x_{i-1},x_i]}f>0. Because of this L\left(P,f,\right)>0. Now consider that since f\in\mathfrak{R} it follows that \int_a^b f~dx=\sup_{P}L\left(P,f\right)>0.

    -------------------------------------------------------------------

    Question 5: Prove that if f\in\mathfrak{R} on [a,b] and c\in[a,b] that \int_a^c f~d\alpha+\int_c^b f~d\alpha=\int_a^b f~d\alpha

    Answer: Let P_1,P_2 be partitions of [a,c],[b,c] respectively. Let P=P_1\cup P_2.

    Let us note that L\left(P,f,\alpha\right)\leq L\left(P_1,f,\alpha\right)+L\left(P_2,f,\alpha\rig  ht). To prove this simply note that given an interval [x_{i-1},x_i] and x\in[x_{i-1},x_i] that (x_{i}-x_{i-1})\inf_{[x_{i-1},x_i]}f\leqslant (x-x_{i-1})\inf_{[x_{i-1},x]}f+(x_i-x)\inf_{[x,x_i]}f.

    So since L\left(P,f,\alpha\right)\leqslant \int_a^c f~d\alpha+\int_c^b f~d\alpha taking the infimum over both intervals gives \int_a^b f~d\alpha\leqslant \int_a^c f~d\alpha+\int_c^b f~d\alpha\quad{\color{red}(1)}.

    Using the same argument except considering that U\left(P_1,f,f\alpha\right)+U\left(P_2,f,\alpha\ri  ght)\leqslant U\left(P,f,\alpha\right) we arrive at \int_a^c f~d\alpha+\int_c^b f~d\alpha\leqslant \int_a^b f~d\alpha\quad{\color{red}(2)}

    Combining \color{red}(1),(2) completes the proof.

    --------------------------------------------------------------------
    Question six: Prove Holder's inequality for integrals, i.e. if \frac{1}{p}+\frac{1}{q}\leqslant 1 then \left|\int_a^b fg~d\alpha\right|\leqslant \left\{\int_a^b f^p~d\alpha\right\}^{\frac{1}{p}}\cdot\left\{\int_  a^b |g|^q~d\alpha\right\}^{\frac{1}{q}}

    I can prove this, but it requires me first to prove that if the conditions of p,q are met as above then uv\leq\frac{u^p}{p}+\frac{v^q}{q} and I am not sure how to do this.

    ----------------------------------------------------------------------

    For 1-5 any criticism is appreciated ...for number six any help with the inequality would be great.

    Thanks in advance
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Mathstud28 View Post

    ------------------------------------------------------------------------
    Question 4: Suppose that f\geqslant 0, f is continuous on [a,b] and \int_a^b f~dx=0. Prove that f(x)=0~~\forall x\in[a,b].

    Answer: It is obvious that if f=0\implies \int_a^b f~dx=0. So instead let us prove that if f>0\implies \int_a^b f~dx>0.

    Since f\ne 0 we can find a Partition of [a,b] such that for some interval [x_{i-1},x_i] we have that f>0~~\forall x \in[x_{i-1},x_i], now since this interval is closed it follows that \inf_{[x_{i-1},x_i]}f>0. Because of this L\left(P,f,\right)>0. Now consider that since f\in\mathfrak{R} it follows that \int_a^b f~dx=\sup_{P}L\left(P,f\right)>0.

    -------------------------------------------------------------------

    For 1-5 any criticism is appreciated ...for number six any help with the inequality would be great.

    Thanks in advance
    hmm, i just wonder about item 4 because i really can't comprehend on the way you did it..
    why prove this:
    Quote Originally Posted by Mathstud28 View Post
    let us prove that if f>0\implies \int_a^b f~dx>0
    whence you want to prove this:
    Quote Originally Posted by Mathstud28 View Post
    Suppose that f\geqslant 0, f is continuous on [a,b] and \int_a^b f~dx=0. Prove that f(x)=0~~\forall x\in[a,b]
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Thank you Kalagota. Let me rephrase this a little better

    Overview: Before I do my proof let me explain what I attempt to do. I will first show that \int_a^b 0~d\alpha=0. I will then show that if f>0 at any point of [a,b] then \int_a^b f~d\alpha>0. This will show that the only function such that \int_a^b f~d\alpha=0 with f\geqslant 0 is f:[a,b]\longmapsto 0.

    Part one: Here I will show that \int_a^b f~d\alpha=0. Consider a partition P of [a,b]. Let [x_{i-1},x_i] be any interval formed by P. It is clear that since f:[x_{i-1},x_i]\longmapsto0 that M_i=\sup_{[x_{i-1},x_i]}f=0. So U\left(P,f,\alpha\right)=\sum_{i=1}^{n}m_i\cdot \Delta\alpha_i=0. So since f:[a,b]\longmapsto0 is continuous it follows that \int_a^b 0~d\alpha=\inf_{P}~U\left(P,f,\alpha\right)=0

    Part two: Let f>0 for some point of [a,b]. We know use the following lemma.

    Lemma: I will show that if f is uniformly continuous and f(c)>0 then there exists a neighborhood of c such that f>0.

    Proof: Since f is uniformly continuous we have that for every \varepsilon>0 there exists a corresponding \delta>0 such that |x-y|<\delta\implies|f(x)-f(y)|<\varepsilon. Now let f(c)>0, this implies that |c-y|<\delta\implies |f(c)-f(y)|<\varepsilon. So assume that there does not exist a neighborhood of c such that f(y)>0, this implies that f(c)<|f(c)-f(y)|. So picking \varepsilon=f(c) we cannot a \delta that satisfies |f(c)-f(y|<\varepsilon which violates continuity.

    So now noticing that f being on continuous on [a,b] implies that f is uniformly continuous on [a,b] since [a,b] is compact, we can see that there exists a [d,e]\subset [a,b] such that f([d,e])>0. So now pick a partition of P such that d,e\in P. It is clear that there is an interval [x_{i-1},x_i]\subset [d,e]. So we can see that m_i=\inf_{[x_{i-1},x_i]}f>0 which implies that L\left(P,f,\alpha\right)>0. So now since f is continuous we can see that \int_a^b f~d\alpha=\sup_{P}L\left(P,f,\alpha\right)>0.

    This proves that the only continuous function f on [a,b] such that \int_a^b f~d\alpha=0 is f:[a,b]\longmapsto0\quad\blacksquare
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  4. #4
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    Opalg's Avatar
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    These all look correct to me. One small point on Q.2:

    Quote Originally Posted by Mathstud28 View Post
    Now following a similar argument noting that \mathbb{Q} is dense in any subset of the reals we can see that for any partition U\left(P,f,\alpha\right)=1.
    The supremum of the function in any interval of the partition is 1, but the upper sum U for the partition is \alpha(b)-\alpha(a).

    Quote Originally Posted by Mathstud28 View Post
    Question six: Prove Holder's inequality for integrals, i.e. if \frac{1}{p}+\frac{1}{q}\leqslant 1 then \left|\int_a^b fg~d\alpha\right|\leqslant \left\{\int_a^b |f|^p~d\alpha\right\}^{\frac{1}{p}}\cdot\left\{\in  t_a^b |g|^q~d\alpha\right\}^{\frac{1}{q}}
    Not true. You need \tfrac1p + \tfrac1q\; {\color{red}=}\; 1 (or possibly {\color{red}\geqslant}\;1).

    Quote Originally Posted by Mathstud28 View Post
    it requires me first to prove that if the conditions of p,q are met as above then uv\leq\frac{u^p}{p}+\frac{v^q}{q} and I am not sure how to do this.
    Assuming that \tfrac1p + \tfrac1q = 1 and that u and v are ≥0, this is elementary calculus. Let f(u,v) = \frac{u^p}{p}+\frac{v^q}{q} - uv. For a fixed value of v, f(u,v) is ≥0 at u=0 and as u→∞. The derivative is 0 when v = u^{p-1}, at which point f has a minimum value of 0.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    The supremum of the function in any interval of the partition is 1, but the upper sum U for the partition is \alpha(b)-\alpha(a).
    Yes, thank you. I had this correct down on my paper

    U\left(P,f,\alpha\right)=\sum_{i=1}^{n}M_i\cdot\De  lta\alpha_i=\sum_{i=1}^{n}\Delta \alpha_i =\left(\alpha(x_1)-\alpha(x_0)\right)+(\alpha(x_2)-\alpha(x_1))+\cdots+(\alpha(x_n)-\alpha(x_{n-1})=\alpha(x_n)-\alpha(x_0) =\alpha(b)-\alpha(a)

    Not true. You need \tfrac1p + \tfrac1q\; {\color{red}=}\; 1 (or possibly {\color{red}\geqslant}\;1).[/tex]
    Pardon me, I transcribed the problem incorrectly.

    Assuming that \tfrac1p + \tfrac1q = 1 and that u and v are ≥0, this is elementary calculus. Let f(u,v) = \frac{u^p}{p}+\frac{v^q}{q} - uv. For a fixed value of v, f(u,v) is ≥0 at u=0 and as u→∞. The derivative is 0 when v = u^{p-1}, at which point f has a minimum value of 0.
    I meant using purely non-calculus means, but I guess this works as well
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  6. #6
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    Quote Originally Posted by Mathstud28 View Post
    Hey everyone! Any input would be appreciated.

    A note on notation, f\in\mathfrak{R}(\alpha) on [a,b] means that f is Riemann-Stieltjes integrable with respect to \alpha(x) on the interval [a,b]

    Question 1: Let x_0\in[a,b] and define f on [a,b] by f:x\longmapsto\left\{ \begin{array}{rcl} 0 & \mbox{if} & x\ne x_0\\ c & \mbox{if} & x=x_0 \end{array} \right.. Also, let \alpha(x) be continuous at x=x_0. Is f\in\mathfrak{R}(\alpha) on [a,b]? If so what is its value.

    Answer: Using the following lemma:

    Lemma: Suppose that f is bounded on [a,b], f has a finite number of discontinuities and \alpha is continuous at all those points, then f\in\mathfrak{R}(\alpha) on [a,b].

    We can see that f\in\mathfrak{R}(\alpha) since f has only one discontinuity, at x_0, and \alpha is continuous there.

    Next to find the value of this integral consider L\left(P,f,\alpha\right) for any partitions. Now consider the interval [x_{i-1},x_i] created by P. Since P is a finite point set, it follows that [x_{i-1},x_i] has infinite points. Which indicates it contains values other than x_0. So for any interval on P we can see that m_i=\inf_{[x_{i-1},x_i]}f=0. Now since L\left(P,f,\alpha\right)=\sum_{i=1}^{n}m_i\Delta \alpha_i we can see that for any partition L\left(P,f,\alpha\right)=0. Now since f\in\mathfrak{R}(\alpha) it follows that \int_a^b f~d\alpha=\inf_{P}L\left(P,f,\alpha\right)=0
    Looks good to me.

    ---------------------------------------------------------------------

    Question 2: Let f:x\longmapsto \left\{ \begin{array}{rcl} 0 & \mbox{if} & x\not\in\mathbb{Q}\\ 1 & \mbox{if} & x\in\mathbb{Q} \end{array} \right., this is known as Dirchlet's function. Show that f\not\in\mathfrak{R}(\alpha) for any a<b

    Answer: This one is simple I think. Consider any partition P of [a,b]. Now consider any interval [x_{i-1},x_i] formed by P. For the same reason as above [x_{i-1},x_i] is infinite. Now since R-\left\{\mathbb{Q}\right\} is dense in any subset of the reals it follows that there are value of x in [x_{i-1},x_i] such that x\not\in\mathbb{Q} where it follows similarly to above that L\left(P,f,\alpha\right)=0 for any partition. Now following a similar argument noting that \mathbb{Q} is dense in any subset of the reals we can see that for any partition U\left(P,f,\alpha\right)=1. So for any partition U\left(P,f,\alpha\right)-L\left(P,f,\alpha\right)=1. Where it follows that f\not\in\mathfrak{R}(\alpha) since it fails the criterion that given any \varepsilon>0 we must have a partition such that U\left(P,f,\alpha\right)-L\left(P,f,\alpha\right)<\varepsilon
    Also good.
    --------------------------------------------------------------------
    Question 3: Does f^2\in\mathfrak{R}(\alpha)\implies f\in\mathfrak{R}(\alpha)?

    Answer: No. Counterexample f:x\longmapsto\left\{ \begin{array}{rcl} -1 & \mbox{if} & x\not\in\mathbb{Q}\\ 1 & \mbox{if} & x\in\mathbb{Q}\end{array}\right.. The reason is that f^2:x\longmapsto\left\{ \begin{array}{rcl} 1 & \mbox{if} & x\not\in\mathbb{Q}\\ 1 & \mbox{if} & x\in\mathbb{Q}\end{array} \right.=1 which is continuous thus integrable, but for the same reason as f in question 2 f is not integrable.
    Good.

    ------------------------------------------------------------------------
    Question 4: Suppose that f\geqslant 0, f is continuous on [a,b] and \int_a^b f~dx=0. Prove that f(x)=0~~\forall x\in[a,b].

    Answer: It is obvious that if f=0\implies \int_a^b f~dx=0.
    Yes, it is obvious. You were also not asked to prove it so there is no need to mention it! (This is NOT an "if and only if" statement.)

    So instead let us prove that if f>0\implies \int_a^b f~dx>0.
    Since f\ne 0 we can find a Partition of [a,b] such that for some interval [x_{i-1},x_i] we have that f>0~~\forall x \in[x_{i-1},x_i], now since this interval is closed it follows that \inf_{[x_{i-1},x_i]}f>0. Because of this L\left(P,f,\right)>0. Now consider that since f\in\mathfrak{R} it follows that \int_a^b f~dx=\sup_{P}L\left(P,f\right)>0.
    Good.
    -------------------------------------------------------------------

    Question 5: Prove that if f\in\mathfrak{R} on [a,b] and c\in[a,b] that \int_a^c f~d\alpha+\int_c^b f~d\alpha=\int_a^b f~d\alpha

    Answer: Let P_1,P_2 be partitions of [a,c],[b,c] respectively. Let P=P_1\cup P_2.

    Let us note that L\left(P,f,\alpha\right)\leq L\left(P_1,f,\alpha\right)+L\left(P_2,f,\alpha\rig  ht). To prove this simply note that given an interval [x_{i-1},x_i] and x\in[x_{i-1},x_i] that (x_{i}-x_{i-1})\inf_{[x_{i-1},x_i]}f\leqslant (x-x_{i-1})\inf_{[x_{i-1},x]}f+(x_i-x)\inf_{[x,x_i]}f.

    So since L\left(P,f,\alpha\right)\leqslant \int_a^c f~d\alpha+\int_c^b f~d\alpha taking the infimum over both intervals gives \int_a^b f~d\alpha\leqslant \int_a^c f~d\alpha+\int_c^b f~d\alpha\quad{\color{red}(1)}.

    Using the same argument except considering that U\left(P_1,f,f\alpha\right)+U\left(P_2,f,\alpha\ri  ght)\leqslant U\left(P,f,\alpha\right) we arrive at \int_a^c f~d\alpha+\int_c^b f~d\alpha\leqslant \int_a^b f~d\alpha\quad{\color{red}(2)}

    Combining \color{red}(1),(2) completes the proof.
    I think you need to say a little more than this. Not every partition of [a, b] can be writtenas P_1\cup P_2 in this way. You need to assert that given any such partition, there exist a refinement that is of that form. If c is not itself a "boundary" of an interval in P, it lies in one of the intervals. Dividing that interval at c gives a refinement that can be written as P_1\cup P_2.

    --------------------------------------------------------------------
    Question six: Prove Holder's inequality for integrals, i.e. if \frac{1}{p}+\frac{1}{q}\leqslant 1 then \left|\int_a^b fg~d\alpha\right|\leqslant \left\{\int_a^b f^p~d\alpha\right\}^{\frac{1}{p}}\cdot\left\{\int_  a^b |g|^q~d\alpha\right\}^{\frac{1}{q}}

    I can prove this, but it requires me first to prove that if the conditions of p,q are met as above then uv\leq\frac{u^p}{p}+\frac{v^q}{q} and I am not sure how to do this.
    That is called "Young's inequality" and a proof can be found here:
    Young's inequality - Wikipedia, the free encyclopedia

    ----------------------------------------------------------------------

    For 1-5 any criticism is appreciated ...for number six any help with the inequality would be great.

    Thanks in advance
    Last edited by HallsofIvy; December 29th 2008 at 07:46 AM.
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    I think you need to say a little more than this. Not every partition of [a, b] can be writtenas P_1\cup P_2 in this way. You need to assert that given any such partition, there exist a refinement that is of that form. If c is not itself a "boundary" of an interval in P, it lies in one of the intervals. Dividing that interval at c gives a refinement that can be written as P_1\cup P_2.
    Thank you HallsofIvy.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Question 1: Let x_0\in[a,b] and define f on [a,b] by f:x\longmapsto\left\{ \begin{array}{rcl} 0 & \mbox{if} & x\ne x_0\\ c & \mbox{if} & x=x_0 \end{array} \right.. Also, let \alpha(x) be continuous at x=x_0. Is f\in\mathfrak{R}(\alpha) on [a,b]? If so what is its value.

    Answer: Using the following lemma:

    Lemma: Suppose that f is bounded on [a,b], f has a finite number of discontinuities and \alpha is continuous at all those points, then f\in\mathfrak{R}(\alpha) on [a,b].

    We can see that f\in\mathfrak{R}(\alpha) since f has only one discontinuity, at x_0, and \alpha is continuous there.

    Next to find the value of this integral consider L\left(P,f,\alpha\right) for any partitions. Now consider the interval [x_{i-1},x_i] created by P. Since P is a finite point set, it follows that [x_{i-1},x_i] has infinite points. Which indicates it contains values other than x_0. So for any interval on P we can see that m_i=\inf_{[x_{i-1},x_i]}f=0. Now since L\left(P,f,\alpha\right)=\sum_{i=1}^{n}m_i\Delta \alpha_i we can see that for any partition L\left(P,f,\alpha\right)=0. Now since f\in\mathfrak{R}(\alpha) it follows that \int_a^b f~d\alpha=\inf_{P}L\left(P,f,\alpha\right)=0
    Would this be a correct alternative solution to the above question? I did not using a lemma in my last one.


    Answer: By definition we know that f\in\mathfrak{R}(\alpha) iff for every \varepsilon>0 there exists a partition P of [a,b] such that U\left(P,f,\alpha\right)-L\left(P,f,\alpha\right)<\varepsilon~~{\color{red}  (*)}. So if we can prove \color{red}(*) for this particular function then we have proved integrability.


    So now consider as I stated in my first solution that since for any subinterval [x_{i-1},x_i] created by the partition P there exists a x\ne x_0 an element of [x_{i-1}x_i]. This shows that m_i=\inf_{[x_{i-1},x_i]}f=0 which in turn implies that L\left(P,f,\alpha\right)=0 for any partition. Thus proving \color{red}(*) reduces to proving U\left(P,f,\alpha\right)<\varepsilon.

    To do this first define a paritition P of [a,b], we can see that M_i=\sup_{[x_{i-1},x_i]}f=\left\{ \begin{array}{rcl} 1 & \mbox{if} & x_0\in[x_{i-1},x_i]\\ 0 & \mbox{if} & x\not\in[x_{i-1},x_i]\end{array}\right.. So now let [x_0-\delta,x_0+\delta] represent the interval centered at x_0, then it is apparent that U\left(P,f,\alpha\right)=\sum_{i=1}^{n}M_i\cdot\De  lta\alpha_i=\alpha\left( x_0+\delta\right)-\alpha\left( x_0-\delta\right). So now we can pick a parition that forces \delta to be as small as we want, this in turn implies that we may pick a partition that makes \alpha\left( x_0+\delta\right)-\alpha\left( x_0-\delta\right) as small as we want. This last fact is due to \alpha being continuous at x_0. This proves that f\in\mathfrak{R}(\alpha)

    So now for the value of the integral I stay with my previous response.



    Im pretty dang sure that is correct...but with analysis I never know . Any criticism is very welcome!
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    So now for the value of the integral I stay with my previous response.
    Sorry to beat a dead horse everyone, I just coming up with things about his basic problem. To compute the value of \int_a^b f~d\alpha couldnt we consider this? Note, let us slightly generliaze this to f:x\longmapsto\left\{\begin{array}{rcl} \xi & \mbox{if} & x=x_0\\ 0 & \mbox{if} & x\ne x_0\end{array}\right.. It is clear that f\in\mathbb{R}(\alpha) following the above arguments.

    So now to find the value first consider that since f\in\mathfrak{R}(\alpha) that for any interval containing x_0 that \int_a^b f~d\alpha=\int_{x_0-\delta}^{x_0+\delta}f~d\alpha+\int_{a}^b f~d\alpha where [c,d] is just "excess" interval not containing x_0. Now Consider that on any interval not containing x_0 \int_a^b f~d\alpha=0, so that \int_a^b f~d\alpha=\int_{x_0-\delta}^{x_0+\delta}f~d\alpha. Now note that repeated reasoning shows that then \int_a^b f~d\alpha=\lim_{\delta\to0}\int_{x_0-\delta}^{x_0+\delta}f~d\alpha. But now consider that since f\geqslant0 that 0\leqslant\int_{x_0-\delta}^{x_0+\delta} f d\alpha\leqslant\xi\left[\alpha_(x_0+\delta)-\alpha(x_0-\delta)\right]. So now then 0\leqslant\int_a^b f~d\alpha=\lim_{\delta\to0}\int_{x_0-\delta}^{x_0+\delta}f~d\alpha\leqslant\lim_{\delta  \to0}\xi\left[\alpha(x_0+\delta)-\alpha(x_0-\delta)\right] and since \alpha is continuous at x_0 this implies that 0\leqslant\int_a^b f~d\alpha\leqslant0\implies\int_a^b f~d\alpha=0

    Just wanted to share.
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