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Since you've answered part of the question, let me elaborate on the Dunford integral. If $f:\Omega\to E$ is weakly measurable and satisfies $\langle x^* ,f\rangle\in L^1(\Omega)$ for all $x^*\in E^\*$ then it's possible to define $\int_X f\in E^{\*\*}$. This acts on $E^{\*}$ via the prescription $$x^*\mapsto \int_X\langle x^{\*},f(\omega)\rangle d\omega.$$ I first learnt this in Talagrand's AMS memoir:

http://books.google.co.uk/books?id=YqJ3t2oT5WgC&lpg=PP1&dq=pettis%20integral%20and%20measure%20theory&pg=PP1#v=onepage&q=pettis%20integral%20and%20measure%20theory&f=false

but a good text on infinite dimensional analysis will probably mention it. $f$ is Pettis integrable precisely when the Dunford integral $\int_Xf$ lies in the canonical image of $E$ in $E^{\*\*}$ and then the two agree (modulo the spaces they live in).

Let's cook up an example. Take the function $f:[0,1]\to c_0$ defined by $f(x)=2^ne_n$ whenever $x\in [2^{-n},2^{-n+1})$. Define $f_n$ to be the obvious simple approximation: let it be $0$ on $[0,2^{-{n+1}})$ and agreeing with $f$ elsewhere.

The problem with these step functions: we compute $$\int_0^1{\| f(s)-f_n(s)\|ds}=\sum_{k=n}^\infty{2^k(2^{-k+1}-2^{-k})}=\infty$$ and note that our dumb approximation doesn't work in the strong sense.

But pick any $g\in\ell^1$ and note that $$\int_0^1{\langle f(s)-f_n(s) , g\rangle ds} =\sum_{k=n}^\infty g(k) \to 0$$ as $n\to\infty$. So our function can be weakly approximated by step functions - in fact you can easily see $s\mapsto \langle f(s),g\rangle$ is $L^1$ with $\int\langle f(s),g\rangle=\lim_{n\to\infty}\int\langle f_n(s),g\rangle$. So $f$ is Dunford integrable.

Note that the integrals $F_n=\int_0^1{f_n(s)ds}=\sum_{k=1}^{n-1}{e_k}$ do not converge to anything in $c_0$. We can think of them as members of $\ell^\infty$ and of course here they converge (in weak* topology) to the identity. In fact, $\int_0^1{\langle f(s), g\rangle ds}=\sum_{k=1}^\infty g(k)$ - so by definition $\int_0^1{f(s)ds}=1 \in \ell^\infty$. This is an example of the Dunford integral, when the Petis/Bochner integrals won't do.

Another example: modify the function $f$ in our example of Dunford integrability, so that $f(x)=2^ne_n/n$ instead of $2^ne_n$. The obvious step function approximation still can't be strong, as $\int_0^1{\|f(s)-f_n(s)\|}=\sum_{k=n}^\infty{1/k}$. Now the sequence of approximations $F_n$ converges in norm to $(1/k)_{k=1}^\infty$. Thought of as an element of $\ell^\infty$ this is the Dunford integral, but as an element of $c_0$ it's just the Pettis integral. (Note, I'm not even saying this function is not Bochner integrable, but the weak approximations make the Pettis integral very natural in this context).

Since you've answered part of the question, let me elaborate on the Dunford integral. If $f:\Omega\to E$ is weakly measurable and satisfies $\langle x^* ,f\rangle\in L^1(\Omega)$ for all $x^*\in E^*$ then it's possible to define $\int_X f\in E^{**}$. This acts on $E^{*}$ via the prescription $$x^*\mapsto \int_X\langle x^{*},f(\omega)\rangle d\omega.$$ I first learnt this in Talagrand's AMS memoir:

http://books.google.co.uk/books?id=YqJ3t2oT5WgC&lpg=PP1&dq=pettis%20integral%20and%20measure%20theory&pg=PP1#v=onepage&q=pettis%20integral%20and%20measure%20theory&f=false

but a good text on infinite dimensional analysis will probably mention it. $f$ is Pettis integrable precisely when the Dunford integral $\int_Xf$ lies in the canonical image of $E$ in $E^{**}$ and then the two agree (modulo the spaces they live in).

Let's cook up an example. Take the function $f:[0,1]\to c_0$ defined by $f(x)=2^ne_n$ whenever $x\in [2^{-n},2^{-n+1})$. Define $f_n$ to be the obvious simple approximation: let it be $0$ on $[0,2^{-{n+1}})$ and agreeing with $f$ elsewhere.

The problem with these step functions: we compute $$\int_0^1{\| f(s)-f_n(s)\|ds}=\sum_{k=n}^\infty{2^k(2^{-k+1}-2^{-k})}=\infty$$ and note that our dumb approximation doesn't work in the strong sense.

But pick any $g\in\ell^1$ and note that $$\int_0^1{\langle f(s)-f_n(s) , g\rangle ds} =\sum_{k=n}^\infty g(k) \to 0$$ as $n\to\infty$. So our function can be weakly approximated by step functions - in fact you can easily see $s\mapsto \langle f(s),g\rangle$ is $L^1$ with $\int\langle f(s),g\rangle=\lim_{n\to\infty}\int\langle f_n(s),g\rangle$. So $f$ is Dunford integrable.

Note that the integrals $F_n=\int_0^1{f_n(s)ds}=\sum_{k=1}^{n-1}{e_k}$ do not converge to anything in $c_0$. We can think of them as members of $\ell^\infty$ and of course here they converge (in weak* topology) to the identity. In fact, $\int_0^1{\langle f(s), g\rangle ds}=\sum_{k=1}^\infty g(k)$ - so by definition $\int_0^1{f(s)ds}=1 \in \ell^\infty$. This is an example of the Dunford integral, when the Petis/Bochner integrals won't do.

Another example: modify the function $f$ in our example of Dunford integrability, so that $f(x)=2^ne_n/n$ instead of $2^ne_n$. The obvious step function approximation still can't be strong, as $\int_0^1{\|f(s)-f_n(s)\|}=\sum_{k=n}^\infty{1/k}$. Now the sequence of approximations $F_n$ converges in norm to $(1/k)_{k=1}^\infty$. Thought of as an element of $\ell^\infty$ this is the Dunford integral, but as an element of $c_0$ it's just the Pettis integral. (Note, I'm not even saying this function is not Bochner integrable, but the weak approximations make the Pettis integral very natural in this context).

Since you've answered part of the question, let me elaborate on the Dunford integral. If $f:\Omega\to E$ is weakly measurable and satisfies $\langle x^* ,f\rangle\in L^1(\Omega)$ for all $x^*\in E^\*$ then it's possible to define $\int_X f\in E^{\*\*}$. This acts on $E^{\*}$ via the prescription $$x^*\mapsto \int_X\langle x^{\*},f(\omega)\rangle d\omega.$$ I first learnt this in Talagrand's AMS memoir:

http://books.google.co.uk/books?id=YqJ3t2oT5WgC&lpg=PP1&dq=pettis%20integral%20and%20measure%20theory&pg=PP1#v=onepage&q=pettis%20integral%20and%20measure%20theory&f=false

but a good text on infinite dimensional analysis will probably mention it. $f$ is Pettis integrable precisely when the Dunford integral $\int_Xf$ lies in the canonical image of $E$ in $E^{\*\*}$ and then the two agree (modulo the spaces they live in).

Let's cook up an example. Take the function $f:[0,1]\to c_0$ defined by $f(x)=2^ne_n$ whenever $x\in [2^{-n},2^{-n+1})$. Define $f_n$ to be the obvious simple approximation: let it be $0$ on $[0,2^{-{n+1}})$ and agreeing with $f$ elsewhere.

The problem with these step functions: we compute $$\int_0^1{\| f(s)-f_n(s)\|ds}=\sum_{k=n}^\infty{2^k(2^{-k+1}-2^{-k})}=\infty$$ and note that our dumb approximation doesn't work in the strong sense.

But pick any $g\in\ell^1$ and note that $$\int_0^1{\langle f(s)-f_n(s) , g\rangle ds} =\sum_{k=n}^\infty g(k) \to 0$$ as $n\to\infty$. So our function can be weakly approximated by step functions - in fact you can easily see $s\mapsto \langle f(s),g\rangle$ is $L^1$ with $\int\langle f(s),g\rangle=\lim_{n\to\infty}\int\langle f_n(s),g\rangle$. So $f$ is Dunford integrable.

Note that the integrals $F_n=\int_0^1{f_n(s)ds}=\sum_{k=1}^{n-1}{e_k}$ do not converge to anything in $c_0$. We can think of them as members of $\ell^\infty$ and of course here they converge (in weak* topology) to the identity. In fact, $\int_0^1{\langle f(s), g\rangle ds}=\sum_{k=1}^\infty g(k)$ - so by definition $\int_0^1{f(s)ds}=1 \in \ell^\infty$. This is an example of the Dunford integral, when the Petis/Bochner integrals won't do.

Another example: modify the function $f$ in our example of Dunford integrability, so that $f(x)=2^ne_n/n$ instead of $2^ne_n$. The obvious step function approximation still can't be strong, as $\int_0^1{\|f(s)-f_n(s)\|}=\sum_{k=n}^\infty{1/k}$. Now the sequence of approximations $F_n$ converges to $(1/k)_{k=1}^\infty$. Thought of as an element of $\ell^\infty$ this is the Dunford integral, but as an element of $c_0$ it's just the Pettis integral. (Note, I'm not even saying this function is not Bochner integrable, but the weak approximations make the Pettis integral very natural in this context).

Since you've answered part of the question, let me elaborate on the Dunford integral. If $f:\Omega\to E$ is weakly measurable and satisfies $\langle x^* ,f\rangle\in L^1(\Omega)$ for all $x^*\in E^\*$ then it's possible to define $\int_X f\in E^{\*\*}$. This acts on $E^{\*}$ via the prescription $$x^*\mapsto \int_X\langle x^{\*},f(\omega)\rangle d\omega.$$ I first learnt this in Talagrand's AMS memoir:

http://books.google.co.uk/books?id=YqJ3t2oT5WgC&lpg=PP1&dq=pettis%20integral%20and%20measure%20theory&pg=PP1#v=onepage&q=pettis%20integral%20and%20measure%20theory&f=false

but a good text on infinite dimensional analysis will probably mention it. $f$ is Pettis integrable precisely when the Dunford integral $\int_Xf$ lies in the canonical image of $E$ in $E^{\*\*}$ and then the two agree (modulo the spaces they live in).

Let's cook up an example. Take the function $f:[0,1]\to c_0$ defined by $f(x)=2^ne_n$ whenever $x\in [2^{-n},2^{-n+1})$. Define $f_n$ to be the obvious simple approximation: let it be $0$ on $[0,2^{-{n+1}})$ and agreeing with $f$ elsewhere.

The problem with these step functions: we compute $$\int_0^1{\| f(s)-f_n(s)\|ds}=\sum_{k=n}^\infty{2^k(2^{-k+1}-2^{-k})}=\infty$$ and note that our dumb approximation doesn't work in the strong sense.

But pick any $g\in\ell^1$ and note that $$\int_0^1{\langle f(s)-f_n(s) , g\rangle ds} =\sum_{k=n}^\infty g(k) \to 0$$ as $n\to\infty$. So our function can be weakly approximated by step functions - in fact you can easily see $s\mapsto \langle f(s),g\rangle$ is $L^1$ with $\int\langle f(s),g\rangle=\lim_{n\to\infty}\int\langle f_n(s),g\rangle$. So $f$ is Dunford integrable.

Note that the integrals $F_n=\int_0^1{f_n(s)ds}=\sum_{k=1}^{n-1}{e_k}$ do not converge to anything in $c_0$. We can think of them as members of $\ell^\infty$ and of course here they converge (in weak* topology) to the identity. In fact, $\int_0^1{\langle f(s), g\rangle ds}=\sum_{k=1}^\infty g(k)$ - so by definition $\int_0^1{f(s)ds}=1 \in \ell^\infty$. This is an example of the Dunford integral, when the Petis/Bochner integrals won't do.

Another example: modify the function $f$ in our example of Dunford integrability, so that $f(x)=2^ne_n/n$ instead of $2^ne_n$. The obvious step function approximation still can't be strong, as $\int_0^1{\|f(s)-f_n(s)\|}=\sum_{k=n}^\infty{1/k}$. Now the sequence of approximations $F_n$ converges in norm to $(1/k)_{k=1}^\infty$. Thought of as an element of $\ell^\infty$ this is the Dunford integral, but as an element of $c_0$ it's just the Pettis integral. (Note, I'm not even saying this function is not Bochner integrable, but the weak approximations make the Pettis integral very natural in this context).

Since you've answered part of the question, let me elaborate on the Dunford integral. If $f:\Omega\to E$ is weakly measurable and satisfies $\langle x^* ,f\rangle\in L^1(\Omega)$ for all $x^*\in E^\*$ then it's possible to define $\int_X f\in E^{\*\*}$. This acts on $E^{\*}$ via the prescription $$x^*\mapsto \int_X\langle x^{\*},f(\omega)\rangle d\omega.$$ I first learnt this in Talagrand's AMS memoir:

http://books.google.co.uk/books?id=YqJ3t2oT5WgC&lpg=PA45&ots=F-Kvn872lZ&dq=dunford%20integral&pg=PP1#v=onepage&q=dunford%20integral&f=false

but a good text on infinite dimensional analysis will probably mention it. $f$ is Pettis integrable precisely when the Dunford integral $\int_Xf$ lies in the canonical image of $E$ in $E^{\*\*}$ and then the two agree (modulo the spaces they live in).

Let's cook up an example. Take the function $f:[0,1]\to c_0$ defined by $f(x)=2^ne_n$ whenever $x\in [2^{-n},2^{-n+1})$. Define $f_n$ to be the obvious simple approximation: let it be $0$ on $[0,2^{-{n+1}})$ and agreeing with $f$ elsewhere.

The problem with these step functions: we compute $$\int_0^1{\| f(s)-f_n(s)\|ds}=\sum_{k=n}^\infty{2^k(2^{-k+1}-2^{-k})}=\infty$$ and note that our dumb approximation doesn't work in the strong sense.

But pick any $g\in\ell^1$ and note that $$\int_0^1{\langle f(s)-f_n(s) , g\rangle ds} =\sum_{k=n}^\infty g(k) \to 0$$ as $n\to\infty$. So our function can be weakly approximated by step functions - in fact you can easily see $s\mapsto \langle f(s),g\rangle$ is $L^1$ with $\int\langle f(s),g\rangle=\lim_{n\to\infty}\int\langle f_n(s),g\rangle$. So $f$ is Dunford integrable.

Note that the integrals $F_n=\int_0^1{f_n(s)ds}=\sum_{k=1}^{n-1}{e_k}$ do not converge to anything in $c_0$. We can think of them as members of $\ell^\infty$ and of course here they converge to the identity. In fact, $\int_0^1{\langle f(s), g\rangle ds}=\sum_{k=1}^\infty g(k)$ - so by definition $\int_0^1{f(s)ds}=1 \in \ell^\infty$. This is an example of the Dunford integral, when the Petis/Bochner integrals won't do.

Another example: modify the function $f$ in our example of Dunford integrability, so that $f(x)=2^ne_n/n$ instead of $2^ne_n$. The obvious step function approximation still can't be strong, as $\int_0^1{\|f(s)-f_n(s)\|}=\sum_{k=n}^\infty{1/k}$. Now the sequence of approximations $F_n$ converges to $(1/k)_{k=1}^\infty$. Thought of as an element of $\ell^\infty$ this is the Dunford integral, but as an element of $c_0$ it's just the Pettis integral. (Note, I'm not even saying this function is not Bochner integrable, but the weak approximations make the Pettis integral very natural in this context).

Since you've answered part of the question, let me elaborate on the Dunford integral. If $f:\Omega\to E$ is weakly measurable and satisfies $\langle x^* ,f\rangle\in L^1(\Omega)$ for all $x^*\in E^\*$ then it's possible to define $\int_X f\in E^{\*\*}$. This acts on $E^{\*}$ via the prescription $$x^*\mapsto \int_X\langle x^{\*},f(\omega)\rangle d\omega.$$ I first learnt this in Talagrand's AMS memoir:

http://books.google.co.uk/books?id=YqJ3t2oT5WgC&lpg=PP1&dq=pettis%20integral%20and%20measure%20theory&pg=PP1#v=onepage&q=pettis%20integral%20and%20measure%20theory&f=false

but a good text on infinite dimensional analysis will probably mention it. $f$ is Pettis integrable precisely when the Dunford integral $\int_Xf$ lies in the canonical image of $E$ in $E^{\*\*}$ and then the two agree (modulo the spaces they live in).

Let's cook up an example. Take the function $f:[0,1]\to c_0$ defined by $f(x)=2^ne_n$ whenever $x\in [2^{-n},2^{-n+1})$. Define $f_n$ to be the obvious simple approximation: let it be $0$ on $[0,2^{-{n+1}})$ and agreeing with $f$ elsewhere.

The problem with these step functions: we compute $$\int_0^1{\| f(s)-f_n(s)\|ds}=\sum_{k=n}^\infty{2^k(2^{-k+1}-2^{-k})}=\infty$$ and note that our dumb approximation doesn't work in the strong sense.

But pick any $g\in\ell^1$ and note that $$\int_0^1{\langle f(s)-f_n(s) , g\rangle ds} =\sum_{k=n}^\infty g(k) \to 0$$ as $n\to\infty$. So our function can be weakly approximated by step functions - in fact you can easily see $s\mapsto \langle f(s),g\rangle$ is $L^1$ with $\int\langle f(s),g\rangle=\lim_{n\to\infty}\int\langle f_n(s),g\rangle$. So $f$ is Dunford integrable.

Note that the integrals $F_n=\int_0^1{f_n(s)ds}=\sum_{k=1}^{n-1}{e_k}$ do not converge to anything in $c_0$. We can think of them as members of $\ell^\infty$ and of course here they converge (in weak* topology) to the identity. In fact, $\int_0^1{\langle f(s), g\rangle ds}=\sum_{k=1}^\infty g(k)$ - so by definition $\int_0^1{f(s)ds}=1 \in \ell^\infty$. This is an example of the Dunford integral, when the Petis/Bochner integrals won't do.

Another example: modify the function $f$ in our example of Dunford integrability, so that $f(x)=2^ne_n/n$ instead of $2^ne_n$. The obvious step function approximation still can't be strong, as $\int_0^1{\|f(s)-f_n(s)\|}=\sum_{k=n}^\infty{1/k}$. Now the sequence of approximations $F_n$ converges to $(1/k)_{k=1}^\infty$. Thought of as an element of $\ell^\infty$ this is the Dunford integral, but as an element of $c_0$ it's just the Pettis integral. (Note, I'm not even saying this function is not Bochner integrable, but the weak approximations make the Pettis integral very natural in this context).