For your question to make sense, you have to actually have an infinite sequence $X_1,X_2,\dots$ iid random vectors. Then, by the strong law of large numbers (SLLN) , $$\frac1n\,\sum_{i=1}^n f(X_i)\to Ef(X_1)=\int_{\mathbb R^p} f\, dp$$ almost surely (a.s.), which, as desired, implies the weak law of large numbers (WLLN) , that is, the convergence in probability (which is always implied by the a.s. convergence).

Concerning your second question, it is of course not true that "when a sequence of random variables converge[s] to a [constant] in probability, the convergence is a.s." -- That would be the same as to say that the convergence in probability always implies the a.s. convergence, because $Y_n\to Y$ a.s. (respectively, in probability) if and only if $Y_n-Y$ converges to the constant $0$ a.s. (respectively, in probability).

For your question to make sense, you have to actually have an infinite sequence $X_1,X_2,\dots$ of iid random vectors. Then, by the strong law of large numbers (SLLN) , $$\frac1n\,\sum_{i=1}^n f(X_i)\to Ef(X_1)=\int_{\mathbb R^p} f\, dp$$ almost surely (a.s.), which, as desired, implies the weak law of large numbers (WLLN) , that is, the convergence in probability (which is always implied by the a.s. convergence).

Concerning your second question, it is of course not true that "when a sequence of random variables converge[s] to a [constant] in probability, the convergence is a.s." -- That would be the same as to say that the convergence in probability always implies the a.s. convergence, because $Y_n\to Y$ a.s. (respectively, in probability) if and only if $Y_n-Y$ converges to the constant $0$ a.s. (respectively, in probability).

For your question to make sense, you have to actually have an infinite sequence $X_1,X_2,\dots$ iid random vectors. Then, by the strong law of large numbers (SLLN), $$\frac1n\,\sum_{i=1}^n f(X_i)\to Ef(X_1)=\int_{\mathbb R^p} f\, dp$$ almost surely, which implies the weak law of large numbers (SLLN), that is, the convergence in probability.

For your question to make sense, you have to actually have an infinite sequence $X_1,X_2,\dots$ iid random vectors. Then, by the strong law of large numbers (SLLN) , $$\frac1n\,\sum_{i=1}^n f(X_i)\to Ef(X_1)=\int_{\mathbb R^p} f\, dp$$ almost surely (a.s.), which, as desired, implies the weak law of large numbers (WLLN) , that is, the convergence in probability (which is always implied by the a.s. convergence).

Concerning your second question, it is of course not true that "when a sequence of random variables converge[s] to a [constant] in probability, the convergence is a.s." -- That would be the same as to say that the convergence in probability always implies the a.s. convergence, because $Y_n\to Y$ a.s. (respectively, in probability) if and only if $Y_n-Y$ converges to the constant $0$ a.s. (respectively, in probability).