Integral of average of integral equal to integral itself











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I was trying to prove the following and had totally no idea how to start.



Let $f$ be integrable over $(-infty, infty)$, and let $h> 0$ be fixed. Prove that $$int_{-infty}^{infty} (frac{1}{2h} int_{x-h}^{x+h} f(y) mathrm{d}y) mathrm{d}x = int_{-infty}^{infty}f(x) mathrm{d}x$$



When $f$ is given integrability, I can use linearity but not so sure if that is the right thing to do here. Also I was considering double integral as the left hand side of the equality looks like that. That is my intention, for example with the help of Fubini's Theorem. However $f$ is defined on one dimension, I struggled with constructing a bi-variate function.






real-analysis lebesgue-integral







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  • Try writing the LHS as an integral from $-R$ to $R$ and then take the limit, usually with these kinds of problems you just have to do the math head on and suddenly the magic happens.
    – tommy1996q
    6 hours ago

















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I was trying to prove the following and had totally no idea how to start.



Let $f$ be integrable over $(-infty, infty)$, and let $h> 0$ be fixed. Prove that $$int_{-infty}^{infty} (frac{1}{2h} int_{x-h}^{x+h} f(y) mathrm{d}y) mathrm{d}x = int_{-infty}^{infty}f(x) mathrm{d}x$$



When $f$ is given integrability, I can use linearity but not so sure if that is the right thing to do here. Also I was considering double integral as the left hand side of the equality looks like that. That is my intention, for example with the help of Fubini's Theorem. However $f$ is defined on one dimension, I struggled with constructing a bi-variate function.






real-analysis lebesgue-integral







share|cite|improve this question






















  • Try writing the LHS as an integral from $-R$ to $R$ and then take the limit, usually with these kinds of problems you just have to do the math head on and suddenly the magic happens.
    – tommy1996q
    6 hours ago















up vote
0
down vote

favorite









up vote
0
down vote

favorite











I was trying to prove the following and had totally no idea how to start.



Let $f$ be integrable over $(-infty, infty)$, and let $h> 0$ be fixed. Prove that $$int_{-infty}^{infty} (frac{1}{2h} int_{x-h}^{x+h} f(y) mathrm{d}y) mathrm{d}x = int_{-infty}^{infty}f(x) mathrm{d}x$$



When $f$ is given integrability, I can use linearity but not so sure if that is the right thing to do here. Also I was considering double integral as the left hand side of the equality looks like that. That is my intention, for example with the help of Fubini's Theorem. However $f$ is defined on one dimension, I struggled with constructing a bi-variate function.






real-analysis lebesgue-integral







share|cite|improve this question













I was trying to prove the following and had totally no idea how to start.



Let $f$ be integrable over $(-infty, infty)$, and let $h> 0$ be fixed. Prove that $$int_{-infty}^{infty} (frac{1}{2h} int_{x-h}^{x+h} f(y) mathrm{d}y) mathrm{d}x = int_{-infty}^{infty}f(x) mathrm{d}x$$



When $f$ is given integrability, I can use linearity but not so sure if that is the right thing to do here. Also I was considering double integral as the left hand side of the equality looks like that. That is my intention, for example with the help of Fubini's Theorem. However $f$ is defined on one dimension, I struggled with constructing a bi-variate function.






real-analysis lebesgue-integral




real-analysis lebesgue-integral






share|cite|improve this question













share|cite|improve this question











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asked 6 hours ago









adosdeci

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276












  • Try writing the LHS as an integral from $-R$ to $R$ and then take the limit, usually with these kinds of problems you just have to do the math head on and suddenly the magic happens.
    – tommy1996q
    6 hours ago




















  • Try writing the LHS as an integral from $-R$ to $R$ and then take the limit, usually with these kinds of problems you just have to do the math head on and suddenly the magic happens.
    – tommy1996q
    6 hours ago


















Try writing the LHS as an integral from $-R$ to $R$ and then take the limit, usually with these kinds of problems you just have to do the math head on and suddenly the magic happens.
– tommy1996q
6 hours ago






Try writing the LHS as an integral from $-R$ to $R$ and then take the limit, usually with these kinds of problems you just have to do the math head on and suddenly the magic happens.
– tommy1996q
6 hours ago












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LHS $=frac 1 {2h}int_{-infty}^{infty} int_{y-h}^{y+h} dx f(y) dy$ by Fubini's Theorem because $x-h<y<x+h$ is equivalent to $ y-h<x<y+h$. Rest is obvious.






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    LHS $=frac 1 {2h}int_{-infty}^{infty} int_{y-h}^{y+h} dx f(y) dy$ by Fubini's Theorem because $x-h<y<x+h$ is equivalent to $ y-h<x<y+h$. Rest is obvious.






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      LHS $=frac 1 {2h}int_{-infty}^{infty} int_{y-h}^{y+h} dx f(y) dy$ by Fubini's Theorem because $x-h<y<x+h$ is equivalent to $ y-h<x<y+h$. Rest is obvious.






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        up vote
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        LHS $=frac 1 {2h}int_{-infty}^{infty} int_{y-h}^{y+h} dx f(y) dy$ by Fubini's Theorem because $x-h<y<x+h$ is equivalent to $ y-h<x<y+h$. Rest is obvious.






        share|cite|improve this answer












        LHS $=frac 1 {2h}int_{-infty}^{infty} int_{y-h}^{y+h} dx f(y) dy$ by Fubini's Theorem because $x-h<y<x+h$ is equivalent to $ y-h<x<y+h$. Rest is obvious.







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        share|cite|improve this answer










        answered 6 hours ago









        Kavi Rama Murthy

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