Is a constrained optimization problem equalivant to its Lagrangian form?

For the following problem:

$\text{min:}\ f(x)\\ s.t. \ g(x)\leq t$

Is the above problem equivalent to the following problem?

$\text{min:}\ f(x) + \lambda g(x) \\ s.t. \ \lambda\geq0$

where $t$ and $\lambda$ are variables. It seems equivalent, because if we increase $\lambda$, $t$ will tend to decrease, if we decrease $\lambda$, then $t$ will tend to increase.

This question is not clear, as you don't define what you're minimizing over, among other things. If you are minimizing with respect to $(x, \lambda)$, the answer is "no", because the minimum (for $g(x) \geq 0$) will occur at $\lambda = 0$ and $x = \arg \min f(x)$, i.e., at the solution to the unconstrained problem $\min_x f(x)$.
However, it is true that, under a broad range of conditions, there exists a $\lambda$ such that the solution to $\min_x:f(x) + \lambda g(x)$ equals the solution to the constrained problem as initially stated. It is not sufficient to constrain $\lambda \geq 0$ to find that solution; instead, you must solve the root-finding problem:
$$\min_\lambda \text{s.t.} \, g(x(\lambda)) - t \leq 0$$
where $x(\lambda) = \arg \min_x f(x) + \lambda g(x)$. ($\text{s.t.}$ means "such that" or "subject to".) The reason for choosing the minimum $\lambda$ from the set of $\lambda$s that cause the constraint to be satisfied is that, writing heuristicaly, as you increase $\lambda$ past the minimum, in "most" problems you'll also move $g(x(\lambda))$ farther and farther below $t$ while increasing $f(x(\lambda))$, i.e., you won't be minimizing $f(x)$ subject to the constraint, but subject to some other constraint $g(x) \leq t'$ with $t' < t$.