TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27

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Continuity under the integral sign

Statement

Let $I \subseteq R$ be an interval and let $f : X \times I \to C$ be such that:

for every $t \in I$ , the function $x \mapsto f (x, t)$ is integrable;
for almost every $x$ , the map $t \mapsto f (x, t)$ is continuous on $I$ ;
there is a nonnegative measurable function $g$ with $\int g d μ < + \infty$ and $∣ f (x, t) ∣ \leq g (x)$ for every $t \in I$ and almost every $x$ .

Then $F (t) : = \int f (x, t) d μ (x)$ is continuous on $I$ .

Facts & Assumptions

Given: An interval $I$ and a function $f$ satisfying the three displayed hypotheses.

[L1]

Dominated convergence applies to integrable complex-valued functions under a single $L^{1}$ majorant (Dominated convergence).

Proof

technique · direct

1.1

Fix $t_{0} \in I$ and let $(t_{n})$ be any sequence in $I$ with $t_{n} \to t_{0}$ . For almost every $x$ , continuity in $t$ gives $f (x, t_{n}) \to f (x, t_{0})$ , and the dominating bound gives $∣ f (x, t_{n}) ∣ \leq g (x)$ .

givenL1

2.1

Apply [L1] to the sequence $x \mapsto f (x, t_{n})$ . Then $F (t_{n}) = \int f (x, t_{n}) d μ (x) ⟶ \int f (x, t_{0}) d μ (x) = F (t_{0}) .$ Since every convergent sequence in $I$ has this property, $F$ is continuous at $t_{0}$ , and therefore on all of $I$ .

step 1.1L1∎

Depends on

Dominated convergence

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Gerald B. Folland, Real Analysis, 2nd ed., Theorem 2.27 (standard reference, not scraped)