Integrable function translation
$begingroup$
For real numbers α< β and γ > 0, show that if g is integrable over [α+γ, β+γ], then
$int_{alpha}^{beta} g(t+gamma)dt = int_{alpha+gamma}^{beta+gamma} g(t)dt$
Prove this change of variables formula by successively considering simple functions, bounded
measurable functions, nonnegative integrable functions, and general integrable functions.
My proof:
Let g be integrable over [α+γ, β+γ], then $g^+$and $g^-$ are nonegative integrable functions. There exist an increasing sequence ${φ_n }$ of nonegative simple functions such that $g^+=lim φ_n $. Now, since $∫_{α+γ}^{β+γ}χ_{[α+γ,β+γ]} (t) =m([α+γ,β+γ])=m([α,β]+γ)=∫_α^βχ_{[α,β]} (t+γ) $ for the given interval, we have that $∫_{α+γ}^{β+γ} φ_n (t)dt =∫_α^βφ_n (t+γ)dt$ for all n. By the Monotone Convergence Theorem , $∫_{α+γ}^{β+γ}g^+ (t)dt = ∫_α^βg^+(t+γ)dt $. Similarry for $g^-$. Thus $∫_α^βg(t+γ)dt = ∫_{α+γ}^{β+γ}g(t)dt$.
Is this correct or it need something else?
real-analysis integration proof-verification
edited Dec 12 '18 at 13:58
GNUSupporter 8964民主女神 地下教會
13.7k72550
asked Dec 12 '18 at 13:17
AligrusAligrus
304
$endgroup$
add a comment |
$begingroup$
For real numbers α< β and γ > 0, show that if g is integrable over [α+γ, β+γ], then
$int_{alpha}^{beta} g(t+gamma)dt = int_{alpha+gamma}^{beta+gamma} g(t)dt$
Prove this change of variables formula by successively considering simple functions, bounded
measurable functions, nonnegative integrable functions, and general integrable functions.
My proof:
Let g be integrable over [α+γ, β+γ], then $g^+$and $g^-$ are nonegative integrable functions. There exist an increasing sequence ${φ_n }$ of nonegative simple functions such that $g^+=lim φ_n $. Now, since $∫_{α+γ}^{β+γ}χ_{[α+γ,β+γ]} (t) =m([α+γ,β+γ])=m([α,β]+γ)=∫_α^βχ_{[α,β]} (t+γ) $ for the given interval, we have that $∫_{α+γ}^{β+γ} φ_n (t)dt =∫_α^βφ_n (t+γ)dt$ for all n. By the Monotone Convergence Theorem , $∫_{α+γ}^{β+γ}g^+ (t)dt = ∫_α^βg^+(t+γ)dt $. Similarry for $g^-$. Thus $∫_α^βg(t+γ)dt = ∫_{α+γ}^{β+γ}g(t)dt$.
Is this correct or it need something else?
real-analysis integration proof-verification
edited Dec 12 '18 at 13:58
GNUSupporter 8964民主女神 地下教會
13.7k72550
asked Dec 12 '18 at 13:17
AligrusAligrus
304
$endgroup$
add a comment |
$begingroup$
For real numbers α< β and γ > 0, show that if g is integrable over [α+γ, β+γ], then
$int_{alpha}^{beta} g(t+gamma)dt = int_{alpha+gamma}^{beta+gamma} g(t)dt$
Prove this change of variables formula by successively considering simple functions, bounded
measurable functions, nonnegative integrable functions, and general integrable functions.
My proof:
Let g be integrable over [α+γ, β+γ], then $g^+$and $g^-$ are nonegative integrable functions. There exist an increasing sequence ${φ_n }$ of nonegative simple functions such that $g^+=lim φ_n $. Now, since $∫_{α+γ}^{β+γ}χ_{[α+γ,β+γ]} (t) =m([α+γ,β+γ])=m([α,β]+γ)=∫_α^βχ_{[α,β]} (t+γ) $ for the given interval, we have that $∫_{α+γ}^{β+γ} φ_n (t)dt =∫_α^βφ_n (t+γ)dt$ for all n. By the Monotone Convergence Theorem , $∫_{α+γ}^{β+γ}g^+ (t)dt = ∫_α^βg^+(t+γ)dt $. Similarry for $g^-$. Thus $∫_α^βg(t+γ)dt = ∫_{α+γ}^{β+γ}g(t)dt$.
Is this correct or it need something else?
real-analysis integration proof-verification
edited Dec 12 '18 at 13:58
GNUSupporter 8964民主女神 地下教會
13.7k72550
asked Dec 12 '18 at 13:17
AligrusAligrus
304
$endgroup$
For real numbers α< β and γ > 0, show that if g is integrable over [α+γ, β+γ], then
$int_{alpha}^{beta} g(t+gamma)dt = int_{alpha+gamma}^{beta+gamma} g(t)dt$
Prove this change of variables formula by successively considering simple functions, bounded
measurable functions, nonnegative integrable functions, and general integrable functions.
My proof:
Let g be integrable over [α+γ, β+γ], then $g^+$and $g^-$ are nonegative integrable functions. There exist an increasing sequence ${φ_n }$ of nonegative simple functions such that $g^+=lim φ_n $. Now, since $∫_{α+γ}^{β+γ}χ_{[α+γ,β+γ]} (t) =m([α+γ,β+γ])=m([α,β]+γ)=∫_α^βχ_{[α,β]} (t+γ) $ for the given interval, we have that $∫_{α+γ}^{β+γ} φ_n (t)dt =∫_α^βφ_n (t+γ)dt$ for all n. By the Monotone Convergence Theorem , $∫_{α+γ}^{β+γ}g^+ (t)dt = ∫_α^βg^+(t+γ)dt $. Similarry for $g^-$. Thus $∫_α^βg(t+γ)dt = ∫_{α+γ}^{β+γ}g(t)dt$.
Is this correct or it need something else?
real-analysis integration proof-verification
real-analysis integration proof-verification
edited Dec 12 '18 at 13:58
GNUSupporter 8964民主女神 地下教會
13.7k72550
asked Dec 12 '18 at 13:17
AligrusAligrus
304
edited Dec 12 '18 at 13:58
GNUSupporter 8964民主女神 地下教會
13.7k72550
asked Dec 12 '18 at 13:17
AligrusAligrus
304
edited Dec 12 '18 at 13:58
GNUSupporter 8964民主女神 地下教會
13.7k72550
edited Dec 12 '18 at 13:58
GNUSupporter 8964民主女神 地下教會
13.7k72550
edited Dec 12 '18 at 13:58
GNUSupporter 8964民主女神 地下教會
13.7k72550
13.7k72550
asked Dec 12 '18 at 13:17
AligrusAligrus
304
asked Dec 12 '18 at 13:17
AligrusAligrus
304
asked Dec 12 '18 at 13:17
AligrusAligrus
304
304
add a comment |
add a comment |
1 Answer
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$begingroup$
You need to prove the equality for $textit{arbitrary}$ indicator functions, then build the general case as you describe. So, let $E$ be measurable, and using latin letters for convenience, (in what follows $Ecap [a+c,b+c]$ may be empty, but that's ok):
$int^{b+c}_{a+c}chi_E(t)dt=m(Ecap[a+c,b+c])=m((E-c)cap[a,b])=int^b_achi_{E-c}(t)dt=int^b_achi_{E}(t+c)dt.$
The rest of your proof now goes through unchanged.
answered Dec 12 '18 at 16:10
MatematletaMatematleta
11.4k2920
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$begingroup$
You need to prove the equality for $textit{arbitrary}$ indicator functions, then build the general case as you describe. So, let $E$ be measurable, and using latin letters for convenience, (in what follows $Ecap [a+c,b+c]$ may be empty, but that's ok):
$int^{b+c}_{a+c}chi_E(t)dt=m(Ecap[a+c,b+c])=m((E-c)cap[a,b])=int^b_achi_{E-c}(t)dt=int^b_achi_{E}(t+c)dt.$
The rest of your proof now goes through unchanged.
answered Dec 12 '18 at 16:10
MatematletaMatematleta
11.4k2920
$endgroup$
add a comment |
$begingroup$
You need to prove the equality for $textit{arbitrary}$ indicator functions, then build the general case as you describe. So, let $E$ be measurable, and using latin letters for convenience, (in what follows $Ecap [a+c,b+c]$ may be empty, but that's ok):
$int^{b+c}_{a+c}chi_E(t)dt=m(Ecap[a+c,b+c])=m((E-c)cap[a,b])=int^b_achi_{E-c}(t)dt=int^b_achi_{E}(t+c)dt.$
The rest of your proof now goes through unchanged.
answered Dec 12 '18 at 16:10
MatematletaMatematleta
11.4k2920
$endgroup$
add a comment |
$begingroup$
You need to prove the equality for $textit{arbitrary}$ indicator functions, then build the general case as you describe. So, let $E$ be measurable, and using latin letters for convenience, (in what follows $Ecap [a+c,b+c]$ may be empty, but that's ok):
$int^{b+c}_{a+c}chi_E(t)dt=m(Ecap[a+c,b+c])=m((E-c)cap[a,b])=int^b_achi_{E-c}(t)dt=int^b_achi_{E}(t+c)dt.$
The rest of your proof now goes through unchanged.
answered Dec 12 '18 at 16:10
MatematletaMatematleta
11.4k2920
$endgroup$
You need to prove the equality for $textit{arbitrary}$ indicator functions, then build the general case as you describe. So, let $E$ be measurable, and using latin letters for convenience, (in what follows $Ecap [a+c,b+c]$ may be empty, but that's ok):
$int^{b+c}_{a+c}chi_E(t)dt=m(Ecap[a+c,b+c])=m((E-c)cap[a,b])=int^b_achi_{E-c}(t)dt=int^b_achi_{E}(t+c)dt.$
The rest of your proof now goes through unchanged.
answered Dec 12 '18 at 16:10
MatematletaMatematleta
11.4k2920
answered Dec 12 '18 at 16:10
MatematletaMatematleta
11.4k2920
answered Dec 12 '18 at 16:10
MatematletaMatematleta
11.4k2920
answered Dec 12 '18 at 16:10
MatematletaMatematleta
11.4k2920
11.4k2920
add a comment |
add a comment |
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