Is it true that $P(A)=1$ if and only if $A=\Omega$? I was wondering if the following statement is true:

$P(A)=1$ if and only if $A=\Omega$.

Of course we know that $P(\Omega)=1$. But if $P(A)=1$, does it necessarily mean $A=\Omega$? Why?
 A: Let $X\sim Unif([0,1])$ (Also denoted by $\mathcal{U}_{[0,1]}$) be the continuous unifrom random variable defined on some Probability space $(\Omega,\mathcal{F},P)$
Then $\displaystyle P(X^{-1}([0,1]\cap\mathbb{R\setminus Q}))= P(X\text{ is irrational})=\int_{[0,1]\cap \mathbb{R\setminus Q}}\mathbf{1}_{[0,1]}\,d\lambda = 1$ . But $\Omega=X^{-1}([0,1])$ .
Also you can prove that $\displaystyle P(X\text{ is rational})=\int_{[0,1]\cap \mathbb{ Q}}\mathbf{1}_{[0,1]}\,d\lambda=\lambda(\mathbb{Q\cap [0,1]})=0$.
Here $\lambda$ denotes the Lebesgue measure on $[0,1]$
Roughly speaking, if you are picking a real number at random from $[0,1]$ then the "Probability" of getting an irrational number is $1$ but the probability of getting a rational number is $0$. This is due to the fact that any countable set has $0$ Lebesgue Measure. But this does not mean that it is impossible to randomly get a rational number , it just means that the "Probability" of getting it is $0$. This is where the rigourous theory of Probability using Measures come in and sorts these problems out.
So remember that unless you are maybe dealing with discrete probability spaces , you cannot assume that "Probability $1$ means sure occurence and Probability $0$ means an impossible event.
A: No. For example, in $[0, 1]$ with the Lebesgue measure, $[0, 1] - \{0\}$ has Lebesgue measure 1, but $[0, 1] - \{0\}$ is not equal to $[0, 1]$.
A: If you throw a dice repeatedly, the probability of eventually throwing a 6 is one, but it is logically possible that a 6 never occurs, for example every roll of the dice could give a 2. So the answer is no.
A: The other answers have provided specific counterexamples, so I will just discuss some intuition. One fundamental idea in probability theory is that measures (and your $P$ is a measure) do not "see" what happens on sets of measure zero.
Theorems often include the abbreviations a.e. ("almost everywhere") or a.s. ("almost surely"), which roughly mean the theorem ignores sets of measure zero. For example, a result may show that $f_1 = f_2$ a.e., which means that the two functions may be unequal in places, but only on sets having measure zero.
Tying this back to your original question, the event $A$ would be said to happen "almost surely", since $P(A) = 1$. Exceptions may exist, in which case $A \neq \Omega$, but those exceptions have measure zero.
