Answer:
Explanation:
Since (fn) converges pointwise to the zero function, we have that for every x ∈ X,
limn→∞ fn(x) = 0.
Now, for each n ∈ N, we have fn ≥ 0 and thus, max{f1, ..., fn} = fn.
Therefore,
∫max{f1, ..., fn} dμ = ∫fn dμ.
Since ∫max{f1, ..., fn} dμ ≤ M < ∞, we have that ∫fn dμ ≤ M for all n ∈ N.
We can use the dominated convergence theorem to show that
limn→∞ ∫fn dμ = ∫limn→∞ fn dμ = 0.
To apply the dominated convergence theorem, we need to find a dominating function that is integrable and dominates fn for all n. Since fn ≥ 0 for all n, we can take g = max{f1, f2, ...} as our dominating function.
Then, we have that g ≥ fn for all n, and
∫g dμ = ∫max{f1, f2, ...} dμ ≤ M < ∞.
Therefore, g is integrable and dominates fn for all n, and we can apply the dominated convergence theorem to conclude that
limn→∞ ∫fn dμ = ∫limn→∞ fn dμ = 0.