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Show that
$$(b-a)cdot[I(0le ale b)-I(ble ale0)]=int_0^b[I(ale x)-I(ale0)]dx,$$
where $I(cdot)$ is the index function.

By definition, this equation can be verified. Does it have an straightforward proof or how can one observe this equation?

I think we can evaluate both sides of equality and prove it. Let us first evaluate the integral on the RHS. consider two cases:
$$begin{align} &1. quad a ge 0 \
&2. quad a<0 \
end{align}$$
For the first case we have $I(ale0) = 0$, so the integral boils down to $int_0^bI(ale x)dx$.
This integral is equal to: $$ int_0^bI(ale x)dx = begin{cases} 0 qquad qquad qquad quad ; a ge b\ int_a^bI(ale x)dx qquad a<bend{cases} = begin{cases} 0 quad qquad ; age b \ (b-a) quad a<b end{cases}$$

For the second case we have $I(ale0) = 1$ which means $$int_0^b[I(ale x)-I(ale0)]dx = int_0^b(I(ale x)-1)dx = left(int_0^bI(ale x)dx right) -b = int_0^bdx-b = 0 $$
Altogether we have $$ int_0^b[I(ale x)-I(ale0)]dx = begin{cases}0 qquad qquad quad age 0 , bge0 , age b\ (b-a) quad ;;;;; age0 , bge 0, a<b\ 0 qquad qquad quad a<0,bge0 end{cases}$$

since in this case $a<0$.

Also we had in mind that $b>0$. If $b<0$ it is similar to the previous state with a little change in conditions:

$$ int_0^b[I(ale x)-I(ale0)]dx = begin{cases}0 qquad qquad quad age 0 , b<0 , \ -(b-a) quad ;; a<0 , b< 0, a > b\ 0 qquad qquad
quad a<0,b<0, b<aend{cases}$$

Please notice that when $a<0,b<0,a>b$ this means $|a|<|b|$ so the second one is obtained as follow: $$int_0^b[I(ale x)-I(ale0)]dx = int_0^b[I(ale x)-1]dx = int_b^0[1-I(ale x)]dx =int_b^a dx = (a-b)=-(b-a) qquad qquad text{when} qquad a<0 , b< 0, a > b $$
Now consider the LHS. for the LHS we have

$$(b-a)cdot[I(0le ale b)-I(ble ale0)] = begin{cases} 0 qquad qquad quad age 0 , b<0 , \ -(b-a) quad ;; a<0 , b< 0, a > b\ 0 qquad qquad
quad a<0,b<0, b<a \ 0 qquad qquad quad age 0 , bge0 , age b\ (b-a) quad ;;;;; age0 , bge 0, a<b\ 0 qquad qquad quad a<0,bge0 end{cases}$$ which are exactly the same we got from the RHS, only a simple inspection is needed. Sorry for exhaustive explanation but i always wanted someone to explain such integrals to me this way not short and intractable solution. Hope this would do.

Write $int_0^b$ as $int_a^b-int_a^0$. Evaluate
$
J:=int_a^b [I(ale x)-I(ale0)]dx$ by arguing two cases:

• If $ale b$, then $I(ale x)=1$ when $xin[a,b]$, so
  $$J=int_a^b[1-I(ale0)]dx=int_a^b I(a>0)dx=(b-a)I(a>0).tag1$$




• If $a>b$, then $int_a^b=-int_b^a$, and $I(ale x)=0$ when $xin[b,a]$, so
  $$J=-int_b^a [I(ale x)-I(ale0)]dx=int_b^aI(ale0)dx=(a-b)I(ale0).tag2$$

Combining these two cases gives
$$
begin{align}J&=(b-a)I(a>0)I(ale b)+(a-b)I(ale0)I(a>b)\
&=(b-a)I(0<ale b)-(b-a)I(b<ale0).tag3
end{align}$$
This agrees with the LHS of the asserted identity up to some fussiness with $<$ vs $le$; formula (3) is correct for the edge case $b>a=0$ whereas the stated LHS is wrong.

This implies the second integral $int_a^0$ must be zero, which can be verified by plugging $b=0$ into (3). (Apply the general case to the special case!)