Showing that Y has a uniform distribution if Y=F(X) where F is the cdf of continuous X

Let $X$ be a random variable with a continuous and strictly increasing c.d.f. function $F$ (so that the quantile function $F^{−1}$ is well-defined). Define a new random variable $Y$ by $Y = F(X)$. Show that $Y$ has a uniform distribution on the interval $[0, 1]$.

My initial thought is that $Y$ is distributed on the interval $[0,1]$ because this is the range of $F$. But how do you show that it is uniform?

Solutions Collecting From Web of "Showing that Y has a uniform distribution if Y=F(X) where F is the cdf of continuous X"

Let $F_Y(y)$ be the CDF of $Y = F(X)$. Then, for any $y \in [0,1]$ we have:

$F_Y(y) = \Pr[Y \le y] = \Pr[F(X) \le y] = \Pr[X \le F^{-1}(y)] = F(F^{-1}(y)) = y$.

What distribution has this CDF?

$$ Prob(Y\leq x)=P(F(X)\leq x)=P(X\leq F^{-1}(x))=x \\ $$
The last equality is from the definition of the quantile function.

Let $y=g(x)$ be a mapping of the random variable $x$ distributed according to $f(x)$. In the mapping $y=g(x)$ you preserve the condition of probability density (namely you counts the same number of events in the respective bins)

$$
h(y)dy=f(x)dx
$$

where h(y) is the probability distribution of $y$

if $h(y)=1$ (uniform distribution) we have

$$
dy=g'(x)dx=f(x)dx
$$

This means that
$$
g(x)=\int f(x)dx
$$

namely the function $g(x)$ that maps the random variable $x$ distributed according $f(x)$, into a random variable $y$ distributed uniformly is his own cumulative distribution function
$\int f(x) dx$.

Since it is shown by JimmyK4542 that the cdf is equal to $y$, differentiating this with respect to $y$ will yield 1, which will be the pdf of $Y$. And a random variable that is uniformly distributed on $[0,1]$ should have a pdf equal to 1 by definition. Therefore, $Y$ is uniformly distributed over $[0,1]$.