Continuity of integral of indicator function
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I am trying to prove the existence of an equilibrium by applying Brouwer's fixed point theorem. In order to invoke this, I of course need my function to be continuous. The only missing step is finding out whether or not this integral is continuous in c on the intervall [0,1]:
$$p_A = int_{0}^{frac{1}{2}} frac{c}{lambda x + (1-lambda)} mathbb{1} (c in [0, lambda x + (1-lambda)]) mathop{}mathrm{ d} F(x),$$
where $x$ follows an atomless, continuous distribution F(x) on [0,1], and $lambda in [0,1]$.
Can somebody help?
real-analysis integration functions continuity fixed-point-theorems
asked Jan 11 at 22:54
user509037user509037
836
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add a comment |
$begingroup$
I am trying to prove the existence of an equilibrium by applying Brouwer's fixed point theorem. In order to invoke this, I of course need my function to be continuous. The only missing step is finding out whether or not this integral is continuous in c on the intervall [0,1]:
$$p_A = int_{0}^{frac{1}{2}} frac{c}{lambda x + (1-lambda)} mathbb{1} (c in [0, lambda x + (1-lambda)]) mathop{}mathrm{ d} F(x),$$
where $x$ follows an atomless, continuous distribution F(x) on [0,1], and $lambda in [0,1]$.
Can somebody help?
real-analysis integration functions continuity fixed-point-theorems
asked Jan 11 at 22:54
user509037user509037
836
$endgroup$
add a comment |
$begingroup$
I am trying to prove the existence of an equilibrium by applying Brouwer's fixed point theorem. In order to invoke this, I of course need my function to be continuous. The only missing step is finding out whether or not this integral is continuous in c on the intervall [0,1]:
$$p_A = int_{0}^{frac{1}{2}} frac{c}{lambda x + (1-lambda)} mathbb{1} (c in [0, lambda x + (1-lambda)]) mathop{}mathrm{ d} F(x),$$
where $x$ follows an atomless, continuous distribution F(x) on [0,1], and $lambda in [0,1]$.
Can somebody help?
real-analysis integration functions continuity fixed-point-theorems
asked Jan 11 at 22:54
user509037user509037
836
$endgroup$
I am trying to prove the existence of an equilibrium by applying Brouwer's fixed point theorem. In order to invoke this, I of course need my function to be continuous. The only missing step is finding out whether or not this integral is continuous in c on the intervall [0,1]:
$$p_A = int_{0}^{frac{1}{2}} frac{c}{lambda x + (1-lambda)} mathbb{1} (c in [0, lambda x + (1-lambda)]) mathop{}mathrm{ d} F(x),$$
where $x$ follows an atomless, continuous distribution F(x) on [0,1], and $lambda in [0,1]$.
Can somebody help?
real-analysis integration functions continuity fixed-point-theorems
real-analysis integration functions continuity fixed-point-theorems
asked Jan 11 at 22:54
user509037user509037
836
asked Jan 11 at 22:54
user509037user509037
836
asked Jan 11 at 22:54
user509037user509037
836
asked Jan 11 at 22:54
user509037user509037
836
asked Jan 11 at 22:54
user509037user509037
836
836
add a comment |
add a comment |
1 Answer
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By DCT it is enough to show that if $c_n to c$ then $frac {c_n} {lambda x+(1-lambda)} I_{0leq {c_n} leq lambda x+(1-lambda)} to frac c {lambda x+(1-lambda)} I_{0leq c leq lambda x+(1-lambda)} $ for almost all $x$. Note that this convergence holds if $0<c<lambda x+(1-lambda)$. Since $c=lambda x+(1-lambda)$ for at most one value of $x$ and $F$ is atomless we have completed the proof when $c>0$. The case $c=0$ is much simpler. [Finiteness of $p_A$ is required. If $lambda <1$ then $lambda x+(1-lambda)geq (1-lambda)$ which makes the finiteness obvious. If $lambda =1$ and $cneq 0$ note that $int_c^{1/2} frac 1 x dF(x) <infty$].
answered Jan 11 at 23:18
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
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1 Answer
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1 Answer
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$begingroup$
By DCT it is enough to show that if $c_n to c$ then $frac {c_n} {lambda x+(1-lambda)} I_{0leq {c_n} leq lambda x+(1-lambda)} to frac c {lambda x+(1-lambda)} I_{0leq c leq lambda x+(1-lambda)} $ for almost all $x$. Note that this convergence holds if $0<c<lambda x+(1-lambda)$. Since $c=lambda x+(1-lambda)$ for at most one value of $x$ and $F$ is atomless we have completed the proof when $c>0$. The case $c=0$ is much simpler. [Finiteness of $p_A$ is required. If $lambda <1$ then $lambda x+(1-lambda)geq (1-lambda)$ which makes the finiteness obvious. If $lambda =1$ and $cneq 0$ note that $int_c^{1/2} frac 1 x dF(x) <infty$].
answered Jan 11 at 23:18
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
$endgroup$
add a comment |
$begingroup$
By DCT it is enough to show that if $c_n to c$ then $frac {c_n} {lambda x+(1-lambda)} I_{0leq {c_n} leq lambda x+(1-lambda)} to frac c {lambda x+(1-lambda)} I_{0leq c leq lambda x+(1-lambda)} $ for almost all $x$. Note that this convergence holds if $0<c<lambda x+(1-lambda)$. Since $c=lambda x+(1-lambda)$ for at most one value of $x$ and $F$ is atomless we have completed the proof when $c>0$. The case $c=0$ is much simpler. [Finiteness of $p_A$ is required. If $lambda <1$ then $lambda x+(1-lambda)geq (1-lambda)$ which makes the finiteness obvious. If $lambda =1$ and $cneq 0$ note that $int_c^{1/2} frac 1 x dF(x) <infty$].
answered Jan 11 at 23:18
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
$endgroup$
add a comment |
$begingroup$
By DCT it is enough to show that if $c_n to c$ then $frac {c_n} {lambda x+(1-lambda)} I_{0leq {c_n} leq lambda x+(1-lambda)} to frac c {lambda x+(1-lambda)} I_{0leq c leq lambda x+(1-lambda)} $ for almost all $x$. Note that this convergence holds if $0<c<lambda x+(1-lambda)$. Since $c=lambda x+(1-lambda)$ for at most one value of $x$ and $F$ is atomless we have completed the proof when $c>0$. The case $c=0$ is much simpler. [Finiteness of $p_A$ is required. If $lambda <1$ then $lambda x+(1-lambda)geq (1-lambda)$ which makes the finiteness obvious. If $lambda =1$ and $cneq 0$ note that $int_c^{1/2} frac 1 x dF(x) <infty$].
answered Jan 11 at 23:18
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
$endgroup$
By DCT it is enough to show that if $c_n to c$ then $frac {c_n} {lambda x+(1-lambda)} I_{0leq {c_n} leq lambda x+(1-lambda)} to frac c {lambda x+(1-lambda)} I_{0leq c leq lambda x+(1-lambda)} $ for almost all $x$. Note that this convergence holds if $0<c<lambda x+(1-lambda)$. Since $c=lambda x+(1-lambda)$ for at most one value of $x$ and $F$ is atomless we have completed the proof when $c>0$. The case $c=0$ is much simpler. [Finiteness of $p_A$ is required. If $lambda <1$ then $lambda x+(1-lambda)geq (1-lambda)$ which makes the finiteness obvious. If $lambda =1$ and $cneq 0$ note that $int_c^{1/2} frac 1 x dF(x) <infty$].
answered Jan 11 at 23:18
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
edited Jan 11 at 23:27
answered Jan 11 at 23:18
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
answered Jan 11 at 23:18
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
answered Jan 11 at 23:18
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
66.3k42867
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