Fuzzy integral definition

[un student]

I'm having problems undestanding the fuzzy integral definition in my
lecture notes. The definition is given by alpha-levels. First let f be
a function from a certain interval to set F which is a set of certain
kind of fuzzy numbers. That is f(t) is a function from R to [0,1].

Let f_a(t) = [f(t)]_a i.e. the alpha levels of f(t)

Fuzzy integral definition (by alpha-levels a)

A_a = \int_a^b f_a(t) dt =
{ \int_a^b g(t) dt |
g(t) \in f_a(t) forall t in [a,b] }

One thing I can't get is what means g(t) \in f_a(t)? How could a
function be a member of a set? This notation remains a total mystery
to me.

I found old exam asking to calculate a fuzzy integral from a given
function. I tried to apply the definition but got nowhere.

Could someone give some light in this definition or point me to source
with more verbose explanations?

[Dmitry A. Kazakov]

Post by un student
I'm having problems undestanding the fuzzy integral definition in my
lecture notes. The definition is given by alpha-levels. First let f be
a function from a certain interval to set F which is a set of certain
kind of fuzzy numbers. That is f(t) is a function from R to [0,1].
Let f_a(t) = [f(t)]_a i.e. the alpha levels of f(t)
Fuzzy integral definition (by alpha-levels a)
A_a = \int_a^b f_a(t) dt =
{ \int_a^b g(t) dt |
g(t) \in f_a(t) forall t in [a,b] }
One thing I can't get is what means g(t) \in f_a(t)? How could a
function be a member of a set? This notation remains a total mystery
to me.

I guess if it was meant to be

<math>
A_\alpha =
\{ \int_a^b g(t) dt |
g(t) \leq f_\alpha(t) \forall t \in [a,b] \}
</math>

I .e. the alpha cut of the fuzzy integral is a set of plain integrals over
[a,b] computed for each function q dominated by the alpha-cut of f.

P.S. Some work is required to show the premises:

1. the integrals of any such q exit
2. alpha-cuts are indeed nested, i.e. comprise a fuzzy subset of R

P.P.S. There are many definitions of fuzzy integrals. Integral Sugeno comes
in mind, etc.

[un student]

Post by Dmitry A. Kazakov

Post by un student
A_a = \int_a^b f_a(t) dt =
{ \int_a^b g(t) dt |
g(t) \in f_a(t) forall t in [a,b] }

<..>

Post by Dmitry A. Kazakov
I guess if it was meant to be
<math>
A_\alpha =
\{ \int_a^b g(t) dt |
g(t) \leq f_\alpha(t) \forall t \in [a,b] \}
</math>
I .e. the alpha cut of the fuzzy integral is a set of plain integrals over
[a,b] computed for each function q dominated by the alpha-cut of f.

I'm not sure if I still understand this. For example isn't alpha-cut a
set? How come a function could be dominated by a set?

Lets say we have a function f(t)(x) defined as

1+t-x, t\leq x \leq t+1
1-t+x, t-1\leq x \leq t \forall t \in R
0 elsewhere

And I'm supposed to calculate the definite integral \int_0^1 f(t) dt
(this is from an old exam). How should I approach this?

Clearly f(t)(x) is "a triangle" with area 1 and t defines how the
midpoint moves along x-axis. Now is f_a(t) = [a-1+t,1+t-a] = [t-(1-a),
t+(1-a)] or should I take f_a(t) as a function of x, i.e.

f_a(t)(x) =
1+t-x, t\leq x \leq t+(1-a)
1-t+x, t-(1-a) \leq x \leq t
0 otherwise

In the latter interpretation the notation g(t)\leq f_a(t) would make
sense, but if f_a(t) is a function of x how could I integrate it over
t..

This turns up to a mess in my hands :(

Post by Dmitry A. Kazakov
1. the integrals of any such q exit
2. alpha-cuts are indeed nested, i.e. comprise a fuzzy subset of R

These are taken for granted. Functions we are working with are
considered to satisfy certain conditions that quarantee the premises.

Post by Dmitry A. Kazakov
P.P.S. There are many definitions of fuzzy integrals. Integral Sugeno comes
in mind, etc.

What the previously presented definition is called? I tried to google
up some resources but couldn't find anything but some research
articles I can't get access to.

[Dmitry A. Kazakov]

Post by un student

Post by Dmitry A. Kazakov

Post by un student
A_a = \int_a^b f_a(t) dt =
{ \int_a^b g(t) dt |
g(t) \in f_a(t) forall t in [a,b] }

<..>

Post by Dmitry A. Kazakov
I guess if it was meant to be
<math>
A_\alpha =
\{ \int_a^b g(t) dt |
g(t) \leq f_\alpha(t) \forall t \in [a,b] \}
</math>
I .e. the alpha cut of the fuzzy integral is a set of plain integrals over
[a,b] computed for each function q dominated by the alpha-cut of f.

I'm not sure if I still understand this. For example isn't alpha-cut a
set?

Yes, it is.

Post by un student
How come a function could be dominated by a set?

f_\alpha(t) is a fuzzy set for any t. You take all real-valued g's such
that for any t the outcome of g is in the support set of f_\alpha(t), i.e.

<math>
f_\alpha(t)(g(t)) \ge 0
</math>

In other words, g may take any value from _\alpha(t). All these g's get
integrated on [a,b]. The results comprise some crisp set of "possible"
integrals. (It should be shown that with alpha increasing, the set get
narrower) Then the integral is proclaimed to have in x the truth value of
maximum of the alpha over sets of possible integrals containing x. (Zadeh's
extension principle).

Post by un student

Post by Dmitry A. Kazakov
P.P.S. There are many definitions of fuzzy integrals. Integral Sugeno comes
in mind, etc.

What the previously presented definition is called? I tried to google
up some resources but couldn't find anything but some research
articles I can't get access to.

I am unsure if that has any name. Unless I am wrong it is merely a
(Lebesque?) integral extended on the case of fuzzy numbers.

Fuzzy numbers obey the inclusion rule. The idea is that you can split a
fuzzy number into a set of nested crisp sets by cutting it at different
levels (alpha). Then for a given alpha you do whatever operation (e.g.
integral) on all members of these crisp sets in all possible combinations
of (here follows some reasoning about set measures and countability, which
we ignore for the sake of clarity (:-)). The obtained crisp set of the
results for given alpha is then handled as I described above in order to
get a fuzzy set = fuzzy number of the extended operation.

[un student]

Post by Dmitry A. Kazakov

Post by Dmitry A. Kazakov
I guess if it was meant to be
<math>
A_\alpha =
\{ \int_a^b g(t) dt |
g(t) \leq f_\alpha(t) \forall t \in [a,b] \}
</math>
I .e. the alpha cut of the fuzzy integral is a set of plain integrals over
[a,b] computed for each function q dominated by the alpha-cut of f.

<..>

Post by Dmitry A. Kazakov
f_\alpha(t) is a fuzzy set for any t. You take all real-valued g's such
that for any t the outcome of g is in the support set of f_\alpha(t), i.e.
<math>
f_\alpha(t)(g(t)) \ge 0
</math>
In other words, g may take any value from _\alpha(t). All these g's get
integrated on [a,b]. The results comprise some crisp set of "possible"
integrals. (It should be shown that with alpha increasing, the set get
narrower) Then the integral is proclaimed to have in x the truth value of
maximum of the alpha over sets of possible integrals containing x. (Zadeh's
extension principle).

I guess my reaction is "mmmkay...". Lets say I have the earlier
defined function:

f(t)(x) =
1+t-x, t\leq x \leq t+1
1-t+x, t-1\leq x \leq t \forall t \in R
0 elsewhere

And I would like to calculate \int_0^1 f(t) dt. Intuitively speking
this is a triangle-like fuzzy number (or set, for that matter) moving
along x-axis. Lets start with

A_0 = \int_0^1 f_0(t) dt = \{ \int_0^1 g(t) dt | g(t)\leq f_0(t)
\forall t \in [0,1]\}.

Now f_a(t) = [t-(1-a), t+(1-a)] and f_0(t) = [t-1, t+1], hence (I'm
cutting details out)

A_0 = \{ \int_0^1 g(t) dt | g(t)\leq f_0(t) \forall t \in [0,1] \}.

Now for any t < -1 we get min \int_0^1 g(t) dt = -1 (let g(t) = f_0
(t)) and max \int g(t) dt = 0 (g(t) == 0). Respectively for t > -1 we
get max being 1 and hence
A_0 = [-1,1].

As a now increases the area between g(t) and x-axis gets smaller, i.e.
absolute value of the integral decreases. From f_a(t) = [t-(1-a), t+(1-
a)] we can conclude that the width of the "triangle" (g(t), that is)
is t+(1-a)-(t-(1-a)) = t+1-a- t+1-a = 2*(1-a) for and height 1-a
giving the area (1-a)^2, hence A_a = [-(1-a)^2, (1-a)^2 ] which gives

F(x) =
1-sqrt(x), 0 \leq x \leq 1
1-sqrt(-x), -1\leq x < 0
0 otherwise

But I have doubts on this result. I'll double check my algebra later.

At least this gets clearer now. Thanks for your help!