Let (X,M,μ) be a measure space and suppose {fn} are non-negative measurable functions decreasing pointwise to f. Suppose also that ∫f1<∞. Then ∫Xf dμ=limn→∞∫Xfn dμ.

Atempt:Since {fn} are decreasing, and converges pointwise to f, then {−fn} is increasing pointwise to f. So by the monotone convergence theorem

∫X−f dμ=limn→∞∫X−fn dμ and so ∫Xf dμ=limn→∞∫Xfn dμ.

**Answer**

The problem is that −fn increases to −f which is not non-negative, so we can’t apply directly to −fn the monotone convergence theorem. But if we take gn:=f1−fn, then {gn} is an increasing sequence of non-negative measurable functions, which converges pointwise to f1−f. Monotone convergence theorem yields:

limn→+∞∫X(f1−fn)dμ=∫Xlimn→+∞(f1−fn)dμ=∫Xf1dμ−∫Xfdμ

so limn→+∞∫Xfndμ=∫Xfdμ.

Note that the fact that there is an integrable function in the sequence is primordial, indeed, if you take X the real line, M its Borel σ-algebra and μ the Lebesgue measure, and fn(x)={1 if x≥n 0 otherwise

the sequence fn decreases to 0 but ∫Rfndλ=+∞ for all n.

**Attribution***Source : Link , Question Author : Kuku , Answer Author : T. Eskin*