Proof of integration-by-substitution (two questions)

Here's a version of the theorem:

$$\int_a^b f(g(x))g'(x)dx=\int_{g(a)}^{g(b)} f(u)du$$ provided that:

1. $f$ is continuous on an interval $I$,

2. $g'$ is continuous on $[a,b]$,

3. $g[a,b]=I$ (i.e. the image of $g$ on $[a,b]$ is $I$),

4. $[a,b]\subseteq I$.

1) Is condition 4 strictly necessary?

2) Is condition 3 correct, or should it be changed to "$g[a,b] \subseteq I$"?

EDIT: I have restated the above theorem more succinctly. $$\int_a^b f(g(x))g'(x)dx=\int_{g(a)}^{g(b)} f(u)du$$ provided that:

1. $f$ is continuous on $g[a,b]$,
2. $g'$ is continuous on $[a,b]$,
3. $[a,b]\subseteq g[a,b]$.
-

You're only assuming that $f$ and $g'$ is continuous on the interval $I$ which could mean that $f$ and $g'$ doesn't behave "nicely" outside of $I$. The extreme case would be that $f$ and $g'$ aren't even defined outside of $I$. So you need condition 4 (where I think you mean $[a,b]\subseteq I$) in order to make sure that the left hand side is well-defined.
Condition 3 could be relaxed to saying that $g([a,b])\subseteq I$, i.e. the image of $[a,b]$ under $g$ is contained in $I$. This is to ensure that the right hand side as well as $f(g(x))$ on the left hand side is well-defined. Again think of the extreme case where $f$ is not defined outside of $I$.
Thanks. I've edited out the "$\subseteq$" typos. About condition 3: since the interval I is arbitrary, do you think that editing "$=$" to become "$\subseteq$" is really a relaxation though (noting that the order of the conditions isn't important)? –  Ryan Feb 26 '13 at 9:48
@Ryan: If the inclusion $g([a,b])\subset I$ is strict, then you could just pick $I'=g([a,b])$ and then $I'$ would clearly satisfy 1.-3., but I don't think you can be sure that 4. holds. But this doesn't really matter, because 4. is only there to ensure that the integrals and composition makes sense. –  Stefan Hansen Feb 26 '13 at 9:53
This is exactly the unsureness I'm having. I can't decide if $=$ or $\subseteq$ would be the "simpler" version or if the $=$ is even correct. Thanks for your help! –  Ryan Feb 26 '13 at 9:56